Sub-array with Given Sum in Java

May 2, 2023

Sub-array with Given Sum

Here, in this page we will write the program for sub-array with given sum in Java. Given an Array of non negative Integers and a number. You need to print all the starting and ending indices of Sub-arrays having their sum equal to the given integer . We will understand this problem in detail with algorithm and code in Java.

Sub array with given sum in Java

Sub-Array With Given Sum in Java

A Sub-array with given sum code in Java is a program that  to a contiguous subsequence of elements within an array that has a sum equal to a specific target value. In other words, it is a consecutive sequence of elements in an array whose sum is equal to a given sum value. The goal is to find such a sub-array within the given array.

Example : arr[ ] = {1, 4, 0, 0, 3, 10, 5}, sum = 7
Output : Sum found between indexes 1 and 4 Sum of elements between indices 1 and 4 is (4 + 0 + 0 + 3) = 7

On this page we will discuss two different methods for this-

  1. Naive Approach
  2. Sliding Window Approach
Sub-array with given sum in Java
  1. Naive approach: The naive approach to find a sub-array with a given sum in Java involves using a nested loop to check all possible sub-arrays of the given array. For each starting index ‘i’, it checks all sub-arrays starting from ‘i’ and ending at ‘j’, and calculates their sum. If the sum matches the given sum, it prints the indices of the sub-array and returns. If no -with the sum is found, it prints a message indicating the absence of such a sub-array.

Algorithm:

  1. Take the input array arr and target sum sum as input.
  2. Initialize curr_sum to 0.
  3. Traverse the array arr from the first element to the last element with a loop variable i ranging from 0 to n-1, where n is the length of the array.
  4. For each i:
    1. Initialize curr_sum as arr[i].
    2. Traverse the array arr from i+1 to n with a loop variable j ranging from i+1 to n-1.
    3. For each j:
      1. Add arr[j] to curr_sum.
      2. If curr_sum is equal to the target sum, print the starting and ending indices of the sub-array as “Subarray found between indexes i and j” and return.
      3. If curr_sum is greater than the target sum or j is equal to n-1, break out of the loop and continue with the next element i.
  5. If no sub-array is found, print “No sub-array found”.
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Java code for Naive Approach

Run 
import java.util.*;

public class Main {
    
    public static void findSubarray(ArrayList< Integer> arr, int sum) {
        int n = arr.size();
        
        for (int i = 0; i < n; i++) {
            int currSum = 0;
            
            for (int j = i; j < n; j++) {
                currSum += arr.get(j);
                
                if (currSum == sum) {
                    System.out.println("Subarray found between indexes " + i + " and " + j);
                    return;
                }
            }
        }
        
        System.out.println("No subarray found");
    }
    
    public static void main(String[] args) {
        ArrayList arr = new ArrayList<>(Arrays.asList(15, 2, 4, 8, 9, 5, 10, 23));
        int sum = 23;
        
        findSubarray(arr, sum);
    }
}

Output

Subarray found between indexes 1 and 4
  1. Sliding Window approach:The code uses a sliding window approach to search for a sub-array with a given sum. The window, represented by the variables ‘start’ and ‘i’, is moved from the left to the right of the array while maintaining a running sum ‘curr_sum’. If the current sum matches the target sum, the indices of the sub-array are printed. If no sub-array with the sum is found, a message indicating the absence of such a sub-array is printed.

Algorithm:

  1. Take the input array arr and target sum sum as input.
  2. Initialize two variables, start and curr_sum, to the first element of the array, i.e., arr[0].
  3. Traverse the array arr from the second element to the last element with a loop variable i ranging from 1 to n-1, where n is the length of the array.
  4. For each i:
    1. Check if curr_sum is greater than the target sum.
    2. If so, increment start by 1 and subtract arr[start-1] from curr_sum until curr_sum is less than or equal to the target sum.
    3. If curr_sum is equal to the target sum, print the starting and ending indices of the subarray as “Subarray found between indexes start and i” and return.
    4. If curr_sum is less than the target sum and i is less than n, add arr[i] to curr_sum and continue with the next element i.
  5. If no subarray is found, print “No subarray found”.

Java Code for sliding window approach

Run 
import java.util.*;

public class Main {
    public static void findSubarray(int[] arr, int n, int sum) {
        int curr_sum = arr[0];
        int start = 0;

        for (int i = 1; i <= n; i++) {
            while (curr_sum > sum && start < i - 1) {
                curr_sum = curr_sum - arr[start];
                start++;
            }

            if (curr_sum == sum) {
                System.out.println("Subarray found between indexes " + start + " and " + (i - 1));
                return;
            }

            if (i < n)
                curr_sum = curr_sum + arr[i];
        }

        System.out.println("No subarray found");
    }

    public static void main(String[] args) {
        int[] arr = { 15, 2, 4, 8, 9, 5, 10, 23 };
        int n = arr.length;
        int sum = 23;

        findSubarray(arr, n, sum);
    }
}

Output

Subarray found between indexes 1 and 4

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