Given an input array find all subarrays with given sum K

Question

Given an input array we can find a single sub-array which sums to K (given) in linear time, by keeping track of sum found so far and the start position. If the current sum b

Answer 1

This problem is very similar to the combination problem solved here: http://introcs.cs.princeton.edu/java/23recursion/Combinations.java.html

Here is my solution:

public static void main(String[] args) {
    int [] input = {-10, 0, 5, 10, 15, 20, 30};
    int expectedSum = 20;

    combination(new SumObj(new int[0]), new SumObj(input), expectedSum);
}

private static void combination(SumObj prefixSumObj, SumObj remainingSumObj, int expectedSum){
    if(prefixSumObj.getSum() == expectedSum){
        System.out.println(Arrays.toString(prefixSumObj.getElements()));
    } 

    for(int i=0; i< remainingSumObj.getElements().length ; i++){
        // prepare new prefix
        int [] newPrefixSumInput = new int[prefixSumObj.getElements().length + 1];
        System.arraycopy(prefixSumObj.getElements(), 0, newPrefixSumInput, 0, prefixSumObj.getElements().length);
        newPrefixSumInput[prefixSumObj.getElements().length] = remainingSumObj.getElements()[i];
        SumObj newPrefixSumObj = new SumObj(newPrefixSumInput);

        // prepare new remaining
        int [] newRemainingSumInput = new int[remainingSumObj.getElements().length - i - 1];            
        System.arraycopy(remainingSumObj.getElements(), i+1, newRemainingSumInput, 0, remainingSumObj.getElements().length - i - 1);
        SumObj newRemainingSumObj = new SumObj(newRemainingSumInput);

        combination(newPrefixSumObj, newRemainingSumObj, expectedSum);
    }
}

private static class SumObj {
    private int[] elements;
    private int sum;

    public SumObj(int[] elements) {
        this.elements = elements;
        this.sum = computeSum();
    }

    public int[] getElements() {
        return elements;
    }

    public int getSum() {
        return sum;
    }

    private int computeSum(){
        int tempSum = 0;
        for(int i=0; i< elements.length; i++){
            tempSum += elements[i];
        }
        return tempSum;
    }
}

Answer 2

Try this code this can work for you:

private static void printSubArrayOfRequiredSum(int[] array, int requiredSum) {
        for (int i = 0; i < array.length; i++) {
            String str = "[ ";
            int sum = 0;
            for (int j = i; j < array.length; j++) {
                sum = sum + array[j];
                str = str + array[j] + ", ";
                if (sum == requiredSum) {
                    System.out.println(" sum : " + sum + " array : " + str
                            + "]");
                    str = "[ ";
                    sum = 0;
                }
            }
        }
    }

Use this method like :

 int array[] = { 3, 5, 6, 9, 14, 8, 2, 12, 7, 7 };
 printSubArrayOfRequiredSum(array, 14);

Output :

sum : 14 array : [ 3, 5, 6, ]
sum : 14 array : [ 14, ]
sum : 14 array : [ 2, 12, ]
sum : 14 array : [ 7, 7, ]

Answer 3

Quadratic Time: O(n2) in worst case.

private static void findSubArray(int[] is, int N) {
    System.out.println("Continuous sub array of " + Arrays.toString(is) + " whose sum is " + N + " is ");
    List<Integer> arry = new ArrayList<>(is.length);
    for (int i = 0; i < is.length; i++) {
        int tempI = is[i];
        arry.add(tempI);
        for (int j = i + 1; j < is.length; j++) {
            if (tempI + is[j] == N) {
                arry.add(is[j]);
                System.out.println(arry);

            } else if (tempI + is[j] < N) {
                arry.add(is[j]);
                tempI = tempI + is[j];
            } else {
                arry.clear();
                break;
            }
        }
    }

}
public static void main(String[] args) {
    findSubArray(new int[] { 42, 15, 12, 8, 6, 32 }, 26);

    findSubArray(new int[] { 12, 5, 31, 13, 21, 8 }, 49);

    findSubArray(new int[] { 15, 51, 7, 81, 5, 11, 25 }, 41);
}

Answer 4

Solution as given by @Evgeny Kluev coded in Java with a little explanation.

public static void main(String[] args) {
    int[] INPUT = {5, 6, 1, -2, -4, 3, 1, 5};
    printSubarrays(INPUT, 5);
}

private static void printSubarrays(int[] input, int k) {
    Map<Integer, List<Integer>> map = new HashMap<Integer, List<Integer>>();
    List<Integer> initial = new ArrayList<Integer>();
    initial.add(-1);
    map.put(0, initial);
    int preSum = 0;

    // Loop across all elements of the array
    for(int i=0; i< input.length; i++) {
        preSum += input[i];
        // If point where sum = (preSum - k) is present, it means that between that 
        // point and this, the sum has to equal k
        if(map.containsKey(preSum - k)) {   // Subarray found 
            List<Integer> startIndices = map.get(preSum - k);
            for(int start : startIndices) {
                System.out.println("Start: "+ (start+1)+ "\tEnd: "+ i);
            }
        }

        List<Integer> newStart = new ArrayList<Integer>();
        if(map.containsKey(preSum)) { 
            newStart = map.get(preSum);
        }
        newStart.add(i);
        map.put(preSum, newStart);
    }
}

Answer 5

Python code with O(n) space and time complexity

from collections import *
k,presum,arr = 5,0,[1,-1,5,-5,5]
start_indices = defaultdict(list)
start_indices[0] = [-1] #this is required for subarrays starting from 0th index
  
for idx,ele in enumerate(arr):
    presum += ele
    for i in start_indices[presum-k]:
        print(arr[i+1:idx+1])
    start_indices[presum] += [idx]

Answer 6

Solution as given by @Evgeny Kluev coded in c++

#include<bits/stdc++.h>
using namespace std;
int c=0;
// Function to print subarray with sum as given sum
void subArraySum(int arr[], int n, int k)
{
   map<int,vector<int>>m1;
   m1[0].push_back(-1);
   int curr_sum=0;
   for(int i=0;i<n;i++){
       curr_sum=curr_sum+arr[i];
       if(m1.find(curr_sum-k)!=m1.end()){
           vector<int>a=m1[curr_sum-k];
           c+=m1[curr_sum-k].size();
           for(int j=0;j<a.size();j++){  // printing all indexes with sum=k
              cout<<a[j]+1<<" "<<i<<endl;
            }
        }
        m1[curr_sum].push_back(i);
    }  
 }
int main()
{
   int arr[] =  {10,2,0,10,0,10};
   int n = sizeof(arr)/sizeof(arr[0]);
   int sum = 10;
   subArraySum(arr, n, sum);
   cout<<c<<endl; //count of subarrays with given sum
   return 0;
}