Smallest subarray from a given Array with sum greater than or equal to K | Set 2

Given an array A[] consisting of N positive 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: arr[] = {3, 1, 7, 1, 2}, K = 11
Output: 3
Explanation:
The smallest subarray with sum ≥ K(= 11) is {3, 1, 7}.

Input: arr[] = {2, 3, 5, 4, 1}, K = 11
Output: 3
Explanation:
The minimum possible subarray is {3, 5, 4}.

Naive and Binary Search Approach: Refer to Smallest subarray from a given Array with sum greater than or equal to K for the simplest approach and the Binary Search based approaches to solve the problem.



Recursive Approach: The simplest approach to solve the problem is to use Recursion. Follow the steps below to solve the problem:

Below is the implementation of the above approach:

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// C++14 program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the smallest subarray
// sum greater than or equal to target
int smallSumSubset(vector<int> data,
                   int target, int maxVal)
{
    int sum = 0;
 
    for(int i : data)
        sum += i;
 
    // Base Case
    if (target <= 0)
        return 0;
 
    // If sum of the array
    // is less than target
    else if (sum < target)
        return maxVal;
 
    // If target is equal to
    // the sum of the array
    else if (sum == target)
        return data.size();
 
    // Required condition
    else if (data[0] >= target)
        return 1;
 
    else if (data[0] < target)
    {
        vector<int> temp;
        for(int i = 1; i < data.size(); i++)
            temp.push_back(data[i]);
             
        return min(smallSumSubset(temp, target,
                                  maxVal),
               1 + smallSumSubset(temp, target -
                               data[0], maxVal));
    }
}
 
// Driver Code
int main()
{
    vector<int> data = { 3, 1, 7, 1, 2 };
    int target = 11;
     
    int val = smallSumSubset(data, target,
                             data.size() + 1);
     
    if (val > data.size())
        cout << -1;
    else
        cout << val;
}    
 
// This code is contributed by mohit kumar 29
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// Java program for the above approach
import java.util.*;
import java.lang.*;
 
class GFG{
 
// Function to find the smallest subarray
// sum greater than or equal to target
static int smallSumSubset(List<Integer> data,
                          int target, int maxVal)
{
    int sum = 0;
 
    for(Integer i : data)
        sum += i;
 
    // Base Case
    if (target <= 0)
        return 0;
 
    // If sum of the array
    // is less than target
    else if (sum < target)
        return maxVal;
 
    // If target is equal to
    // the sum of the array
    else if (sum == target)
        return data.size();
 
    // Required condition
    else if (data.get(0) >= target)
        return 1;
 
    else if (data.get(0) < target)
    {
        List<Integer> temp = new ArrayList<>();
        for(int i = 1; i < data.size(); i++)
            temp.add(data.get(i));
             
        return Math.min(smallSumSubset(temp, target,
                                             maxVal),
                    1 + smallSumSubset(temp, target -
                                data.get(0), maxVal));
    }
    return -1;
}
     
// Driver Code
public static void main (String[] args)
{
    List<Integer> data = Arrays.asList(3, 1, 7, 1, 2);
    int target = 11;
     
    int val = smallSumSubset(data, target,
                             data.size() + 1);
     
    if (val > data.size())
        System.out.println(-1);
    else
        System.out.println(val);
}
}
 
// This code is contributed by offbeat
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# Python3 program for the above approach
 
# Function to find the smallest subarray
# sum greater than or equal to target
def smallSumSubset(data, target, maxVal):
    # base condition
 
    # Base Case
    if target <= 0:
        return 0
 
    # If sum of the array
    # is less than target
    elif sum(data) < target:
        return maxVal
 
    # If target is equal to
    # the sum of the array
    elif sum(data) == target:
        return len(data)
 
    # Required condition
    elif data[0] >= target:
        return 1
 
    elif data[0] < target:
        return min(smallSumSubset(data[1:], \
        target, maxVal),
                1 + smallSumSubset(data[1:], \
                target-data[0], maxVal))
 
 
# Driver Code
data = [3, 1, 7, 1, 2]
target = 11
 
val = smallSumSubset(data, target, len(data)+1)
 
if val > len(data):
    print(-1)
else:
    print(val)
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// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to find the smallest subarray
// sum greater than or equal to target
static int smallSumSubset(List<int> data,
                   int target, int maxVal)
{
    int sum = 0;
   
    foreach(int i in data)
        sum += i;
   
    // Base Case
    if (target <= 0)
        return 0;
   
    // If sum of the array
    // is less than target
    else if (sum < target)
        return maxVal;
   
    // If target is equal to
    // the sum of the array
    else if (sum == target)
        return data.Count;
   
    // Required condition
    else if (data[0] >= target)
        return 1;
   
    else if (data[0] < target)
    {
        List<int> temp = new List<int>();
        for(int i = 1; i < data.Count; i++)
            temp.Add(data[i]); 
               
        return Math.Min(smallSumSubset(temp, target, 
                                       maxVal), 
                    1 + smallSumSubset(temp, target - 
                                    data[0], maxVal));
    }
    return 0;
}
 
// Driver code
static void Main()
{
    List<int> data = new List<int>();
    data.Add(3);
    data.Add(1);
    data.Add(7);
    data.Add(1);
    data.Add(2);
     
    int target = 11;
       
    int val = smallSumSubset(data, target,
                             data.Count + 1);
       
    if (val > data.Count)
        Console.Write(-1);
    else
        Console.Write(val);
}
}
 
// This code is contributed by divyeshrabadiya07
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Output: 
3

Time Complexity: O(2N)
Auxiliary Space: O(N)

Efficient Approach: The above approach can be optimized using Dynamic programming by memoizing the subproblems to avoid recomputation.

Below is the implementation of the above approach:

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// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function to find the smallest subarray
// with sum greater than or equal target
static int minlt(int[] arr, int target, int n)
{
     
    // DP table to store the
    // computed subproblems
    int[][] dp = new int[arr.length + 1][target + 1];
     
    for(int[] row : dp)
        Arrays.fill(row, -1);
     
    for(int i = 0; i < arr.length + 1; i++)
         
        // Initialize first
        // column with 0
        dp[i][0] = 0;
         
    for(int j = 0; j < target + 1; j++)
 
        // Initialize first
        // row with 0
        dp[0][j] = Integer.MAX_VALUE;
         
    for(int i = 1; i <= arr.length; i++)
    {
        for(int j = 1; j <= target; j++)
        {
             
            // Check for invalid condition
            if (arr[i - 1] > j)
            {
                dp[i][j] = dp[i - 1][j];
            }
            else
            {
                 
                // Fill up the dp table
                dp[i][j] = Math.min(dp[i - 1][j],
                        (dp[i][j - arr[i - 1]]) !=
                        Integer.MAX_VALUE ?
                        (dp[i][j - arr[i - 1]] + 1) :
                        Integer.MAX_VALUE);
            }
        }
    }
 
    // Print the minimum length
    if (dp[arr.length][target] == Integer.MAX_VALUE)
    {
        return -1;
    }
    else
    {
        return dp[arr.length][target];
    }
}
 
// Driver code
public static void main (String[] args)
{
    int[] arr = { 2, 3, 5, 4, 1 };
    int target = 11;
    int n = arr.length;
     
    System.out.print(minlt(arr, target, n));
}
}
 
// This code is contributed by offbeat
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# Python3 program for the above approach
 
import sys
 
# Function to find the smallest subarray
# with sum greater than or equal target
def minlt(arr, target, n):
 
    # DP table to store the
    # computed subproblems
    dp = [[-1 for _ in range(target + 1)]\
    for _ in range(len(arr)+1)]
     
    for i in range(len(arr)+1):
         
        # Initialize first
        # column with 0
        dp[i][0] = 0
         
    for j in range(target + 1):
 
        # Initialize first
        # row with 0
        dp[0][j] = sys.maxsize
 
    for i in range(1, len(arr)+1):
        for j in range(1, target + 1):
 
            # Check for invalid condition
            if arr[i-1] > j:
                dp[i][j] = dp[i-1][j]
 
            else:
                # Fill up the dp table
                dp[i][j] = min(dp[i-1][j], \
                1 + dp[i][j-arr[i-1]])
                 
    return dp[-1][-1]
 
    # Print the minimum length
    if dp[-1][-1] == sys.maxsize:
        return(-1)
    else:
        return dp[-1][-1]
 
# Driver Code
arr = [2, 3, 5, 4, 1]
target = 11
n = len(arr)
 
print(minlt(arr, target, n))
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Output
3

Time Complexity: O(N2)
Auxiliary Space: O(N)

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