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In Java, Array is an object. It is a non-primitive data type which stores values of similar data types.

As per the problem statement we have to find a cumulative sum array which means array elements will be updated with the sum of the current element and all previous elements.

To find the cumulative sum, the array must contain all numeric values.

In this article, you will see how to find the cumulative sum of an array by using Java programming language. Let’s explore.

To Show You Some Instances Instance-1 Suppose the original array is {2, 9, 3, 5, 1, 6}

After updating the array with cumulative sum −

Cumulative sum is [2, 11, 14, 19, 20, 26] Instance-2 Suppose the original array is {2, 9, 11, 5, 15, 6, 10}

After updating the array with cumulative sum −

Cumulative sum is [2, 11, 22, 27, 42, 48, 58] Instance-3 Suppose the original array is {44, 5, 9, 15, 31, 22, 19, 48}

After updating the array with cumulative sum −

Cumulative sum is [44, 49, 58, 73, 104, 126, 145, 193] Algorithm

Step 1 − Declare and initialize an integer array. Also declare and initialize an int variable say ‘sum’ as 0.

Step 2 − Traverse through the array.

Step 3 − Calculate the sum value as, sum = sum + array[i]

Step 4 − Replace the sum value with array[i]

Step 5 − Print the elements of the array.

Syntax

To get the length of an array (number of elements in that array), there is an inbuilt property of array i.e length.

Below refers to the syntax of it −

array.length

where, ‘array’ refers to the array reference.

Multiple Approaches

We have provided the solution in different approaches.

By Using Static Initialization of Array and for Loop

By Using Static Initialization of Array and while Loop

By Using User Defined Method

Let’s see the program along with its output one by one.

Approach-1: By Using Static Initialization of Array and for Loop Example

In this approach, array elements will be initialized in the program. Then as per the algorithm, find the cumulative sum of an array. Here, we have used a loop for traversing array.

import java.util.*; public class Main{ public static void main(String[] args) { int sum = 0; int numbers[] = { 44, 5 , 9, 15, 31, 22, 19, 48 }; for (int i = 0; i < numbers.length; i++) { sum += numbers[i]; numbers[i] = sum; } System.out.println("Cumulative sum is "); System.out.println(Arrays.toString(numbers)); } } Output Cumulative sum is [44, 49, 58, 73, 104, 126, 145, 193] Approach-2: By Using Static Initialization of Array and while Loop Example

In this approach, array elements will be initialized in the program. Then as per the algorithm, find the cumulative sum of an array. Here, we have used a while loop for traversing array.

import java.util.*; public class Main{ public static void main(String[] args){ int sum = 0; int numbers[] = { 44, 5 , 9, 15, 31, 22, 19, 48 }; int i =0; while(i<numbers.length){ sum += numbers[i]; numbers[i] = sum; i++; } System.out.println("Cumulative sum is "); System.out.println(Arrays.toString(numbers)); } } Output Cumulative sum is [44, 49, 58, 73, 104, 126, 145, 193] Approach-3: By Using User Defined Method Example

In this approach, array elements will be initialized in the program. Then call a user defined method by passing the array as parameter and inside method as per the algorithm to find the cumulative sum of an array.

import java.util.*; public class Main{ public static void main(String[] args){ cumulativeSum(); } public static void cumulativeSum(){ int sum = 0; int numbers[] = { 2, 9, 11, 5, 15, 6, 10 }; for (int i = 0; i < numbers.length; i++) { sum += numbers[i]; numbers[i] = sum; } System.out.println("Cumulative sum is "); System.out.println(Arrays.toString(numbers)); } } Output Cumulative sum is [2, 11, 22, 27, 42, 48, 58]

In this article, we explored how to find cumulative sum array by using Java programming language.

You're reading How To Find A Cumulative Sum Array In Java?

2D Array Sorting In Java

Introduction to 2D Array Sorting in Java

The following article provides an outline for 2D Array Sorting in Java. An array of arrays can be a two-dimensional array. The matrices that make up the 2D array represent a collection of rows and columns. Because the elements of 2D arrays can get accessed at random, we can access the individual cells in a 2D array using their indexes, just like we can with one-dimensional arrays.

In a 2D array, a cell has two indexes one is its row number, and the other is its column number. Sorting is a technique for arranging elements in a 2D array in a specific order. The 2D array can be in either ascending or descending order. Let’s see how to sort different ways the 2D array in Java in ascending and descending order.

Examples of 2D Array Sorting in Java

Different examples are as below:

Example #1

Example for 2D array sorting in Java to sort all elements of a 2D Array.

Code:

package jex; import java.util.*; public class demo { public static void sort(int arr[][]) { int i, j, temp; int n=arr.length; int m=arr[0].length; for (i = 0; i < n * m - 1; ++i) { for (j = 0; j < n * m - 1 - i; ++j) { temp = arr[(j + 1) / m][(j + 1) % m]; arr[(j + 1) / m][(j + 1) % m] = arr[j / m][j % m]; arr[j / m][j % m] = temp; } } } } public static void print(int arr[][]) { int i, j; int n=arr.length; int m=arr[0].length; for (i = 0; i < n; ++i) { for (j = 0; j < m; ++j) { System.out.print(arr[i][j]+" "); } System.out.println(); } } public static void main(String[] args) { Scanner sc=new Scanner(System.in); int[][] arr={ { 5, 12, 17, 12, 23}, { 1, 2, 4, 6, 8}, {21, 14, 7, 19, 27}, { 3, 18, 9, 15, 25} }; System.out.println("Array Before Sorting is : "); print(arr); sort(arr); System.out.println("Array After Sorting is : "); print(arr); } }

Output:

As in the above program, the sort() method is useful to iterate each element of a 2D array, and when the current element is greater than the next element, then swap the numbers. Finally, the print method displays all the elements of the 2D array. In the main function, the 2D array is created and printed after and before calling the sort() function, as shown in the above output.

Example #2

Example for 2D array sorting in Java to sort all elements of a 2D array by column-wise.

Code:

package jex; import java.util.*; public class demo { public static void sort(int arr[][]) { int i, j,k, temp; int n=arr.length; int m=arr[0].length; for (k = 0; k < m; ++k) { for (i = 0; i < n; ++i) { for (j = 0; j < n - 1 - i; ++j) { temp = arr[j + 1][k]; arr[j + 1][k] = arr[j][k]; arr[j][k] = temp; } } } } } public static void print(int arr[][]) { int i, j; int n=arr.length; int m=arr[0].length; for (i = 0; i < n; ++i) { for (j = 0; j < m; ++j) { System.out.print(arr[i][j]+" "); } System.out.println(); } } public static void main(String[] args) { Scanner sc=new Scanner(System.in); int[][] arr={ { 5, 12, 17, 12, 23}, { 1, 2, 4, 6, 8}, {21, 14, 7, 19, 27}, { 3, 18, 9, 15, 25} }; System.out.println("Array Before Sorting is : "); print(arr); sort(arr); System.out.println("Array After Sorting is : "); print(arr); } }

Output:

As in the above rewrite program, the sort() method is useful to iterate each element of a 2D array and sort the array column-wise. Finally, the print method displays all the elements of the 2D array. In the main function, the 2D array is created and printed after and before calling the sort() function, as shown in the above output.

Example #3

To sort all elements of a 2D array by row-wise.

Code:

package jex; import java.util.*; public class demo { public static void sort(int arr[][]) { int i, j,k, temp; int n=arr.length; int m=arr[0].length; for(k=0;k<n;++k){ for(i=0;i<m;++i){ for(j=0;j<m-1-i;++j){ temp = arr[k][j+1]; arr[k][j+1] = arr[k][j]; arr[k][j] = temp; } } } } } public static void print(int arr[][]) { int i, j; int n=arr.length; int m=arr[0].length; for (i = 0; i < n; ++i) { for (j = 0; j < m; ++j) { System.out.print(arr[i][j]+" "); } System.out.println(); } } public static void main(String[] args) { Scanner sc=new Scanner(System.in); int[][] arr={ { 5, 12, 17, 12, 23}, { 1, 2, 4, 6, 8}, {21, 14, 7, 19, 27}, { 3, 18, 9, 15, 25} }; System.out.println("Array Before Sorting is : "); print(arr); sort(arr); System.out.println("Array After Sorting is : "); print(arr); } }

Output:

As in the above rewrite program, the sort() method is useful to iterate each element of a 2D array and sort the array row-wise. Finally, the print method displays all the elements of the 2D array. In the main function, the 2D array is created and printed after and before calling the sort() function, as shown in the above output.

Conclusion

Sorting is a technique for arranging elements in a 2D array in a specific order. For example, in a 2D array, a cell has two indexes: its row number and its column number.

Recommended Articles

This is a guide to 2D Array Sorting in Java. Here we discuss the introduction and examples of 2D array sorting in Java, respectively. You may also have a look at the following articles to learn more –

Java Program To Count Inversions Of Size Three In A Given Array

Inversion count is a step counting method by which we can calculate the number of sorting steps taken by a particular array. It is also capable to count the operation time span for an array. But, if we want to sort an array in a reverse manner, the count will be maximum number present in that array.

Array: { 5, 4, 3, 2, 1} Pairs: {5, 4}, {5,3} , {3,2}, {3,1}, {2,1},{4,3}, {4,2}, {4,1},}, {5,2}, {5,1} Output: 10 Array: {1, 2, 3, 4, 5} Pairs: No Pairs Output: 0 Array: {1,5,2,8,3,4} Pairs: {5, 2}, {5, 3}, {5, 4}, {8, 3}, {8, 4} Output: 5

The inversion count indicates that how far that particular array is from being sorted in an increasing order. Here are two particular process to describe this situation attached with a solution −

To find the smaller elements − To find out the smaller element from an array, we need to iterate the index from n-1 to 0. By applying (a[i]-1), we can calculate the getSum() here. The process will run until it reach to a[i]-1.

To find the greater number − To find the greater number from an index we need to perform iteration 0 to n-1. For the every element we need to do calculation for every number till a[i]. Subtract it from i. Then we will get a the number which is greater than a[i].

Algorithm to count inversions of size three in an array:-

Here in this algorithm; we learn how to count inversions of size three in a given array in a particular programming environment.

Step 1 − Start

Step 3 − Inner loop y=x+1 to N

Step 4 − If element at x is greater than element at y index

Step 5 − Then, increase the invCount++

Step 6 − Print the pair

Step 7 − Terminate

Syntax to count inversions of size three in an array:-

C++ Implementation

int getInversions(int * A, int n) { int count = 0; for (int i = 0; i < n; ++i) { for (int j = i + 1; j < n; ++j) { ++count; } } } return count; }

Java Implementation

public static int getInversions(int[] A, int n) { int count = 0; for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { count += 1; } } } return count; }

Python Implementation

def getInversions(A, n): count = 0 for i in range(n): for j in range(i + 1, n): count += 1 return count; }

Here we have mentioned the possible syntaxes to count inversions of size three in a given array. And for this method; Time Complexity is O(N^2), where N is the total size of the array and; Space Complexity:O(1), as no extra space has been used.

Approaches to follow

Approach 1 − Count Inversions of size three in a given array by program to count inversions of size 3

Approach 2 − Better Approach to count inversions of size 3

Approach 3 − Count inversions of size 3 using binary indexed tree

Count Inversions of size three in a given array by program to count inversions of size 3

For the simple approach to count inversions of size three, we need to run a loop for all possible value of i, j and k. The time complexity is O(n^3) and O(1) reflects the auxiliary space.

The condition is:

Example 1 public class Inversion{ int getInvCount(int arr[], int n){ int invcount = 0; for(int i=0 ; i< n-2; i++){ for(int j=i+1; j<n-1; j++){ for(int k=j+1; k<n; k++){ invcount++; } } } } return invcount; } public static void main(String args[]){ Inversion inversion = new Inversion(); int arr[] = new int[] {8, 4, 2, 1}; int n = arr.length; System.out.print("Inversion count after method: " + inversion.getInvCount(arr, n)); } } Output Inversion count after method: 4 A better approach to count inversions of size 3

In this method we will consider the every element of an array as middle element of inversion. It helps to reduce the complexity. For this approach, the time complexity is O(n^2) and auxiliary Space is O(1).

Example 2 public class Inversion { int getInvCount(int arr[], int n){ int invcount = 0; for (int i=0 ; i< n-1; i++){ int small=0; for (int j=i+1; j<n; j++) small++; int great = 0; if (arr[i] < arr[j]) great++; invcount += great*small; } return invcount; } public static void main(String args[]){ Inversion inversion = new Inversion(); int arr[] = new int[] {8, 4, 2, 1}; int n = arr.length; System.out.print("Inversion count afret the operation : " + inversion.getInvCount(arr, n)); } } Output Inversion count afret the operation : 4 Count inversions of size 3 using binary indexed tree

In this method, we count the greater elements and smaller ones too. Then perform the multiply operation greater[] to smaller[] and add it to the final result. Here the time complexity is O(n*log(n)) and auxiliary space denoted as O(n).

Example 3 import java.io.*; import java.util.Arrays; import java.util.ArrayList; import java.lang.*; import java.util.Collections; public class rudrabytp { static int N = 100005; static int BIT[][] = new int[4][N]; static void updateBIT(int t, int i, int val, int n){ while (i <= n) { BIT[t][i] = BIT[t][i] + val; i = i + (i & (-i)); } } static int getSum(int t, int i){ int res = 0; res = res + BIT[t][i]; i = i - (i & (-i)); } return res; } static void convert(int arr[], int n){ int temp[]=new int[n]; for (int i = 0; i < n; i++) temp[i] = arr[i]; Arrays.sort(temp); for (int i = 0; i < n; i++) { arr[i] = Arrays.binarySearch(temp,arr[i]) + 1; } } public static int getInvCount(int arr[], int n){ convert(arr, n); updateBIT(1, arr[i], 1, n); for (int l = 1; l < 3; l++) { updateBIT(l + 1, arr[i], getSum(l, arr[i] - 1), n); } } return getSum(3, n); } public static void main (String[] args){ int arr[] = {8, 4, 2, 1}; int n = arr.length; System.out.print("Inversion Count After The Operation : "+getInvCount(arr, n)); } } Output Inversion Count After The Operation : 4 Conclusion

In this article, we have learnt today how to count inversions of size three in a given array. Hope with this article and the mentioned codes using the particular language, you have got a broad view about this topic.

Golang Program To Find Maximum Sum Of A Subarray With Length K

In this article we are going to understand how to use methods naming brute-force, sliding window and prefix sum methods of golang to find the maximum sum of a subarray with length k. We will also discuss the Algorithm for each method and provide code Examples to demonstrate their implementation.

Syntax func len(v Type) int

The len() function is used to get the length of a any parameter. It takes one parameter as the data type variable whose length we wish to find and returns the integer value which is the length of the variable.

Example 1

The first Example to find maximum sum of a subarray with length k is Brute-Force Method.

package main import ( "fmt" "math" ) func maxSumBruteForce(arr []int, k int) int { n := len(arr) maxSum := math.MinInt64 for i := 0; i <= n-k; i++ { sum := 0 for j := i; j < i+k; j++ { sum += arr[j] } maxSum = sum } } return maxSum } func main() { x := []int{1, 2, 3, 4, 5} fmt.Println("The given array of integers is:", x) var num int = 5 result := maxSumBruteForce(x, num) fmt.Println("The max sum is:", result) } Output The given array of integers is: [1 2 3 4 5] The max sum is: 15 Example 2

In this Example we will write a go language program to find the maximum sum of a subarray having k number of elements by using prefix sum method.

package main import ( "fmt" "math" ) func maxSumPrefixSum(arr []int, k int) int { n := len(arr) maxSum := math.MinInt64 prefixSum := make([]int, n+1) for i := 1; i <= n; i++ { prefixSum[i] = prefixSum[i-1] + arr[i-1] } for i := k; i <= n; i++ { sum := prefixSum[i] - prefixSum[i-k] maxSum = sum } } return maxSum } func main() { x := []int{1, 2, 3, 4, 5, 6} fmt.Println("The given array of integers is:", x) var num int = 6 result := maxSumPrefixSum(x, num) fmt.Println("The max sum is:", result) } Output The given array of integers is: [1 2 3 4 5 6] The max sum is: 21 Conclusion

In this article, we explored three different methods to find the maximum sum of a subarray of a given length k in Golang. Here we have used two methods viz. The brute-force method and prefix sum method. The brute force method has a time complexity of O(nk), while prefix sum methods have a time complexity of O(n). The prefix sum methods is more efficient than the brute-force method, as it avoids unnecessary calculations.

Find Maximum Element In Each Row Of A Matrix In Java

In Java, Array is an object. It is a non-primitive data type which stores values of similar data type. The matrix in java is nothing but a multi-dimensional array which represents multiple rows and columns.

Here we have given a matrix which contains set of elements and as per the problem statement we have to find out the maximum element in each row of that matrix.

Let’s deep dive into this article, to know how it can be done by using Java programming language.

To show you some instances 21 22 23 24 25 26 27 28 29

Maximum element in each row: 23, 26 and 29

121 222 243 432 124 245 256 657 237 258 229 345 176 453 756 343

Maximum element in each row: 432, 657, 345 and 756

1 2 3 4 5 6 7 8 9

Maximum element in each row: 3, 6 and 9

Algorithm Algorithm-1: (By Using for Loop)

Step-1 − Use two nested for loops to iterate through the matrix and find the maximum element in each row.

Step-2 − The outer loop iterates through each row of the matrix, while the inner loop iterates through each element in the row.

Step-3 − The maximum element in the row is found by using the chúng tôi method, which returns the larger of two values.

Step-4 − The result is stored in an array, which is returned at the end.

Algorithm-2: (Using Java Stream)

Step-1 − Use Java’s stream API to find the maximum element in each row.

Step-2 − The IntStream.range method is used to create a stream of integers from 0 to the number of rows in the matrix.

Step-3 − The map method is used to apply the function IntStream.of(matrix[i]).max().getAsInt() to each integer in the stream.

Step-4 − This function returns the maximum element in the row matrix[i] as an integer.

Step-5 − The result is stored in an array, which is returned at the end.

Syntax

The Matrix.length() method in Java returns the length of the given matrix.

Below refers to the syntax of it −

inputMatrix.lenght

where, ‘inputMatrix’ refers to the given matrix.

IntStream.range() is a method in Java that generates a sequential stream of integers from startInclusive to endExclusive – 1.

IntStream.range(startInclusive, endExclusive)

It can be used to perform operations on a set of integers. For example, in the second code, it is used to loop through the rows and columns of the matrix.

Multiple Approaches

We have provided the solution in different approaches.

By Using for Loop

By Using Stream Method

Let’s see the program along with its output one by one.

Approach-1: By Using Nested for Loop

In this approach, matrix elements will be initialized in the program. Then calls a user defined method by passing the matrix as parameter and inside method as per the algorithm using a for loop find maximum element in each row of that matrix.

Example import java.util.Arrays; public class Main { public static void main(String[] args) { int[][] inputMatrix = {{12, 12, 3}, {74, 65, 64}, {57, 28, 49}}; System.out.println("Max element in each row: "); int[] result = maxInEachRow(inputMatrix); System.out.println(Arrays.toString(result)); } public static int[] maxInEachRow(int[][] mat) { int[] result = new int[mat.length]; for (int i = 0; i < mat.length; i++) { int max = Integer.MIN_VALUE; for (int j = 0; j < mat[i].length; j++) { max = Math.max(max, mat[i][j]); } result[i] = max; } return result; } } Output Max element in each row: [12, 74, 57] Approach-2: By Using Stream Method

In this approach, matrix elements will be initialized in the program. Then calls a user defined method by passing the matrix as parameter and inside method as per the algorithm using stream method find maximum element in each row of that matrix.

Example import java.util.Arrays; import java.util.stream.IntStream; public class Main { public static void main(String[] args) { int[][] matrix = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}; System.out.println("Max element in each row: "); int[] result = maxInEachRow(matrix); System.out.println(Arrays.toString(result)); } public static int[] maxInEachRow(int[][] matrix){ return IntStream.range(0, matrix.length) .toArray(); } } Output Max element in each row: [3, 6, 9]

In this article, we explored different approaches to find the maximum element in each row of the matrix by using Java programming language.

C++ Program To Find The Gcds Of Given Index Ranges In An Array

In the field of data structure, a range query is a pre-processing method to operate on some input data in an efficient manner. A range query is responsible to answer any query of the particular input on any data subset. If we want to copy some data columns from a table we need to maintain an index for that particular dataset. An index is a direct link or a key, which is designed to provide an efficient searching process in a data set. It is mainly used to speed up the data retrieving from a lost data source.

In mathematics, Greatest Common Divisor aka GCD is a largest possible integer which can divide both of the integers present as the input. Here, all numbers must be present with a non-zero value. Just take an example:

GCD of 70, 80 = 10 (10 is the largest number which divides them with remainder as 0) GCD of 42, 120, 285 = 3 (3 is the largest number which divides them with remainder as 0) Algorithm to find the GCDs of given index ranges in an array (in Detailed)

Step 1 − Start

Step 2 − Construct a section of arr[0] to arr[n-1]

Step 3 − Continue the equal partition

Step 4 − Recursive call for these two parts

Step 5 − For each, save only the greatest common divisor value will save in a segment tree

Step 6 − Build another segment tree to fill it from bottom to top

Step 7 − Each node stores some data of GCD with a certain range

Step 8 − If the node range is startQuery and endQuery, then return a value node

Step 9 − Else if, the range is invalid, it will return a null or -1 as output

Step 10 − Else, return a GCD function as a recursive call

Step 11 − Terminate

Algorithm to find the GCDs of given index ranges in an array (in Short) 

Step 1 − Assume, a and b are the two non-zero integers

Step 2 − Let, a mod b = R

Step 3 − If, a=b and b=R

Step 4 − Then, repeat step 2 and step 3

Step 5 − Process will run until a mod b become greater than zero

Step 6 − GCD = b

Step 7 − Terminate

Syntax to find the GCDs of given index ranges in an array Begin if c = 0 OR d = 0, then return 0 if c = d, then return b return findGCD(c-d, d) else return findGCD(c, d-c) End

Here in this syntax we can see the possible logic code, how to find the Greatest Common Divisor of two non-zero digits. The time complexity for the process is O(Q*N*log(Ai)) and the auxiliary space is evaluated as O(1).

Approach to follow:-

Approach 1 − Program to find GCD of a number in a given Range using segment Trees

Program to find GCD of a number in a given Range using segment Trees

To find GCD of a number in a given Range using segment Trees, we need to follow some unavoidable steps.

Construction of a segment tree:

The elements of an input array are the leaf nodes.

Each individual internal node represents the GCD of all leaf nodes.

Array representation can be done by a segment tree.

-2*(i+1), index’s left element

-2*(i+2), index’s right element

-Parent is floor((i-1)/2)

Construction of a new segment tree by using the given array:

Begin the process with a segment arr[0 . . . n-1].

Divide them into two halves.

Call same for the both halves.

Store the value of GCD.

Construction of given range for GCD:

For every possible query, move the halves of thee tree present left and right.

When the given range overlaps on a half; return the node.

When it lies outside the given range, return 0 at that moment.

For partial overlapping, traverse and get return according the method follows.

Example

using namespace std; int* st; int findGcd(int ss, int se, int qs, int qe, int si) { return 0; return st[si]; int mid = ss + (se – ss) / 2; return __gcd(findGcd(ss, mid, qs, qe, si * 2 + 1), findGcd(mid + 1, se, qs, qe, si * 2 + 2)); } int findRangeGcd(int ss, int se, int arr[], int n) { cout << “Invalid Arguments” << “n”; return -1; } return findGcd(0, n – 1, ss, se, 0); } int constructST(int arr[], int ss, int se, int si) { if (ss == se) { st[si] = arr[ss]; return st[si]; } int mid = ss + (se – ss) / 2; st[si] = __gcd(constructST(arr, ss, mid, si * 2 + 1), constructST(arr, mid + 1, se, si * 2 + 2)); return st[si]; } int* constructSegmentTree(int arr[], int n) { int height = (int)(ceil(log2(n))); int size = 2 * (int)pow(2, height) – 1; st = new int[size]; constructST(arr, 0, n – 1, 0); return st; } int main() { int a[] = { 20, 30, 60, 90, 50 }; int n = sizeof(a) / sizeof(a[0]); constructSegmentTree(a, n); int l = 1; int r = 3; cout << “GCD of the given range is here. Please collect your data:”; cout << findRangeGcd(l, r, a, n) << “n”;

return 0; }

Output GCD of the given range is here. Please collect your data:30 Conclusion

In this article thus, we have developed some possible code by using the particular programming environment. With these encoded logic and the mentioned algorithm we have learned how to find out the GCDs of given index ranges in an array properly.

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