Smallest subarray from a given Array with sum greater than or equal to K

Given an array A[] consisting of N integers and an integer K, the task is to find the length of the smallest subarray with sum greater than or equal to K. If no such subarray exists, print -1.

Examples:

Input: A[] = {2, -1, 2}, K = 3
Output: 3
Explanation:
Sum of the given array is 3.
Hence, the smallest possible subarray satisfying the required condition is the entire array.
Therefore, the length is 3.

Input: A[] = {2, 1, 1, -4, 3, 1, -1, 2}, K = 5
Output: 4

Naive Approach:
The simplest approach to solve the problem is to generate all possible subarrays of the given array and check which subarray sum is greater than or equal to K. Among all such subarrays satisfying the condition, print the subarray having minimum length.
Time Complexity:O(N2)
Auxiliary Space: O(1)



Efficient Approach:
The above approach can be further optimized using Prefix Sum Array and Binary search. Follow the steps below:

  • Initialize an array to store the Prefix sum of the original array.
  • Hash the prefix sum array with the indices using a Map.
  • If a greater sum with a lesser index is already found, then there is no point of hashing a prefix sum which is smaller than the largest prefix sum obtained till now. Therefore, hash the increasing order of prefix sum.
  • Traversing the array and if any element is greater than or equal to K, return 1 as the answer.
  • Otherwise, for every element, perform Binary Search over the indices (i, n-1) in the prefix sum array to find the first index with sum at least K.
  • Return the minimum length subarray obtained from the above steps.

Below is the implementation of the above approach:

C++

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// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to perform Binary Search
// and return the smallest index with
// sum greater than value
int binary_search(map<int, vector<int> >& m,
                  int value, int index)
{
    // Search the value in map
    auto it = m.lower_bound(value);
  
    // If all keys in the map
    // are less then value
    if (it == m.end())
        return 0;
  
    // Check if the sum is found
    // at a greater index
    auto it1
        = lower_bound(it->second.begin(),
                      it->second.end(), index);
  
    if ((it1 - it->second.begin())
        != it->second.size())
        return *it1;
  
    return 0;
}
  
// Function to find the smallest subarray
// with sum greater than equal to K
int findSubarray(int arr[], int n, int k)
{
  
    // Prefix sum array
    int pre_array[n];
  
    // Stores the hashes to prefix sum
    map<int, vector<int> > m;
  
    pre_array[0] = arr[0];
    m[pre_array[0]].push_back(0);
  
    // If any array element is
    // greater than equal to k
    if (arr[0] >= k)
        return 1;
  
    int ans = INT_MAX;
    for (int i = 1; i < n; i++) {
  
        pre_array[i]
            = arr[i] + pre_array[i - 1];
  
        // If prefix sum exceeds K
        if (pre_array[i] >= k)
  
            // Update size of subarray
            ans = min(ans, i + 1);
  
        auto it = m.rbegin();
  
        // Hash prefix sum in
        // increasing order
        if (pre_array[i] >= it->first)
            m[pre_array[i]].push_back(i);
    }
  
    for (int i = 1; i < n; i++) {
  
        int temp
            = binary_search(m,
                            pre_array[i - 1] + k,
                            i);
        if (temp == 0)
            continue;
  
        // Update size of subarray
        ans = min(ans, temp - i + 1);
    }
  
    // If any subarray is found
    if (ans <= n)
        return ans;
  
    // If no such subarray exists
    return -1;
}
  
// Driver Code
int main()
{
    int arr[] = { 2, 1, 1, -4, 3, 1, -1, 2 };
  
    int k = 5;
  
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << findSubarray(arr, n, k) << endl;
  
    return 0;
}

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Output:

4

Time Complexity: O(NlogN)
Auxiliary Space: O(N)

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