Show Buttons
Share On Facebook
Share On Twitter
Share On Google Plus
Share On Linkdin
Share On Pinterest
Share On Reddit
Share On Stumbleupon
Contact us
Hide Buttons

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

You may also like...

  • 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.