Find Minimum Length Sub Array With Sum K

Given an array A having positive and negative integers and a number k, find the minimum length sub array of A with sum = k.


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Algorithm/Insights

1. Iterate over the array using 2 loops.
2. Initialize currentSum = 0, min = MAX_VALUE
3. Starting from array[i], keep adding array[i] to currentSum till currentSum != k or till last element of the array or size of current subarray becomes > min.
4. If currentSum == k update min.
5. Also keep track of start and end index of the min subarray obtained so far.
6. Print array elements from start to end.


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Algorithm Visualization




Code Snippet

			
package com.ideserve.saurabh.questions;

/**
 * <b>IDeserve <br>
 * <a href="https://www.youtube.com/c/IDeserve">https://www.youtube.com/c/IDeserve</a>
 * <br><br>
 * Minimum Length Subarray with Sum = k</b><br>
 * Given an array A having positive and negative integers and a number k, 
 * find the minimum length sub array of A with sum = k.
 * <br><br>
 * Example: <br>
 * array = { 2, 3, 1, 2, 4, 3 }
 * <br>
 * k = 7
 * <br>
 * Output: [ 4 3 ]
 * <br><br>
 * <a href="https://www.youtube.com/watch?v=gHSoIwnERF0">Minimum Length Subarray with Sum k - Youtube Link</a> 
 * @author Saurabh
 *
 */
public class MinimumLengthSubArraySum {

	public static void main(String[] args) {
		int[] array = {2,3,1,2,4,3};
		int k = 7;
		printMinSubArrayWithSum(array, k);
	}
	
	public static void printMinSubArrayWithSum(int[] array, int k) {
		 
	    int start = -1;		// Start of min subarray
	    int end = -1;		// End of min subarray
		int min = Integer.MAX_VALUE;	// size of the smallest subarray with sum = k
		
	    for(int i = 0; i < array.length; i++) {
	    	
	        int currentSum = 0;
	        for(int j = i; j < array.length && (j-i+1) < min; j++) {        
	            currentSum += array[j];
	            if(currentSum == k) {
	                start = i;
	                end = j;
	                min = end - start + 1;
	                break;
	            }            
	        }    
	    }

	    if(start == -1 || end == -1)  {
	    	System.out.println("No subarray exists with sum = " + k);
	        return ;
	    }

	    System.out.print("[ ");
	    while(start <= end) {
	        System.out.print(array[start] + " ");
	        start++;;
	    }		
	    System.out.println("]");
	}

}
		

Order of the Algorithm

Time Complexity is O(n^2)
Space Complexity is O(1)


Contribution

  • Sincere thanks from IDeserve community to Saurabh Kumar for compiling current post.

    Saurabh Kumar

    Ninja Programmer