In an Array, find the Contiguous Subarray with Sum to a Given Value.

Objective: Given an array and an integer, find the Subarray whose sum is equal to the given integer.


int[] arrA = { 25, 12, 14, 22, 19, 15, 10, 23 };
Integer = 55
Output : 55 is found between indexes 2 and 4
And Elements are : 14 22 19

Approach :

Naive Approach: Use 2 loops . Time Complexity – O(n2).

Better Approach: Time Complexity – O(n)

  • Idea is to keep adding all the elements to the currSum
  • Keep checking if the currSum<Sum
  • If currSum gets greater than the Sum then start reducing the currSum from the beginning of the array using “start”if any time currSum=Sum, Stop and return

NOTE: Technique used in this problem can be used to solve the problem “Find a subarray such that the sum of elements in it is equal to zero“. In that the array will aldo have to contain the negative elements and our given sum is zero.

Complete Code:


55 is found between indexes 2 and 4
And Elements are :  14 22 19

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If you find anything incorrect or you feel that there is any better approach to solve the above problem, please write comment.

  • Deep Shah

    I think it can be done in O(n) time using Kadane’s Algorithm (

    • tutorialhorizon

      Yes it can be, but the solution provided is also O(n).

      • Deep Shah

        tutorialhorizon Thanks for the reply.

        This is the other solution which I thought of . It uses only one for loop. and no nested loops inside.

        // Kadanes Algorithm

        private static int maxSubarray(int[] arr) {

        // TODO Auto-generated method stub

        int max_sum = 0;

        int sum = 0;

        int startIndex = 0; // intially start index is at 0

        int endIndex = 0; // intially end index is at 0 position

        int tempStart = 0; // intially start = tempstart = 0

        if(arr.length == 0){

        System.out.println(“No Elements in Array”);

        return -1;


        else if(arr.length == 1){

        System.out.println(“Array contains only single element”);

        startIndex = 0;

        endIndex = 0;

        System.out.println(“Start Index : “+startIndex + “t End Index :”+endIndex);

        return arr[0];


        for(int i = 0; i<arr.length;i++){

        sum = sum + arr[i];

        if(max_sum < sum){

        max_sum = sum;

        startIndex = tempStart; // max_sum < sum so we found a greater sum from last tempstart to current element

        endIndex = i; // make end as current element


        else if(sum < 0){

        sum = 0;

        tempStart = i + 1; // if sum is less than 0, make a fresh start with right element



        System.out.println("Start Index : "+startIndex + "t End Index :"+endIndex);

        System.out.print("Largest Sum is obtained from : ");

        for(int i = startIndex; i<= endIndex;i++){

        System.out.print(arr[i]+", ");


        return max_sum;


        • Holden

          Have you solved this problem with Kandane’s algorithm? Could you please write your code in Ideone and share its link here?
          Since it is not readable here.

  • Holden

    Thank you for the great post.

    Don’t you think this ‘for’:

    for (int i = 0; i <= arrA.length; i++) {

    should be:

    for (int i = 0; i < arrA.length; i++) {


  • Holden

    There is a while in ‘for’ loop, is it still O(n) solution? Your proposed logic seems linear; But in textbooks, we read if there is two nested loop, time complexity will be quadratic.

    • Holden

      I think I got it! These 2 loops are independent! If two loops are not independent, time complexity will be O(n^2). Thank you

      • tutorialhorizon

        Yes holden you got it right. Though there are two nested loops but still every element will be traversed only once.

  • jafar basha

    In an Array, find the Smallest Contiguous Subarray with Sum to a Given Value.

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