Data Modelling & AI Data Structure & Algorithm

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

23 June 2025

3

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 a 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:

If K ? 0: Print -1 as no such subarray can be obtained.
If the sum of the array is equal to K, print the length of the array as the required answer.
If the first element in the array is greater than K, then print 1 as the required answer.
Otherwise, proceed to find the smallest subarray both by considering the current element in the subarray as well as not including it.
Repeat the above steps for every element traversed.

Below is the implementation of the above approach:

C++14

// 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

Java

// 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

Python3

# 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)

C#

// 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

Javascript

<script>
 
// js program for the above approach
 
// Function to find the smallest subarray
// sum greater than or equal to target
function smallSumSubset(data, target, maxVal)
{
    let sum = 0;
 
    for(let i=0;i< data.length;i++) 
        sum += data[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.length;
 
    // Required condition
    else if (data[0] >= target)
        return 1;
 
    else if (data[0] < target)
    {
        let temp = [];
        for(let i = 1; i < data.length; i++)
            temp.push(data[i]); 
             
        return Math.min(smallSumSubset(temp, target, 
                                  maxVal), 
               1 + smallSumSubset(temp, target - 
                               data[0], maxVal));
    }
}
 
// Driver Code
let data = [ 3, 1, 7, 1, 2 ];
let target = 11;
let val = smallSumSubset(data, target,
                             data.length + 1);
     
    if (val > data.length)
        document.write( -1);
    else
        document.write(val);
  
</script>

Output

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

Efficient Approach: The above approach can be optimized using Dynamic programming by memorizing the subproblems to avoid re-computation.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
     
// Function to find the smallest subarray
// with sum greater than or equal target
int minlt(vector<int> arr, int target, int n)
{
     
    // DP table to store the
    // computed subproblems
    vector<vector<int>> dp(arr.size() + 1 ,
           vector<int> (target + 1, -1)); 
     
    for(int i = 0; i < arr.size() + 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] = INT_MAX;
         
    for(int i = 1; i <= arr.size(); 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] = min(dp[i - 1][j],
                   (dp[i][j - arr[i - 1]]) !=
                   INT_MAX ? 
                   (dp[i][j - arr[i - 1]] + 1) : 
                   INT_MAX);
            }
        }
    }
 
    // Print the minimum length
    if (dp[arr.size()][target] == INT_MAX)
    {
        return -1;
    }
    else
    {
        return dp[arr.size()][target]; 
    }
}
 
// Driver code
int main()
{
    vector<int> arr = { 2, 3, 5, 4, 1 };
    int target = 11;
    int n = arr.size();
     
    cout << minlt(arr, target, n);
}
 
// This code is contributed by Surendra_Gangwar

Java

// 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

Python3

# 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))

C#

// C# program for the
// above approach
using System;
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 i = 0;
          i < arr.Length + 1; i++)
  {
    for (int j = 0; 
             j < target + 1; j++) 
    {
      dp[i, j] = -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] = int.MaxValue;
 
  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]]) !=
                           int.MaxValue ? 
                           (dp[i, j - 
                            arr[i - 1]] + 1) : 
                           int.MaxValue);
      }
    }
  }
 
  // Print the minimum 
  // length
  if (dp[arr.Length, 
         target] == int.MaxValue)
  {
    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;
  Console.Write(
  minlt(arr, target, n));
}
}
 
// This code is contributed by gauravrajput1

Javascript

<script>
 
// JavaScript program for the above approach
     
// Function to find the smallest subarray
// with sum greater than or equal target
function minlt(arr, target, n)
{
     
    // DP table to store the
    // computed subproblems
    var dp = Array.from(Array(arr.length+1), 
    ()=>Array(target+1).fill(-1));
     
    for(var i = 0; i < arr.length + 1; i++)
     
        // Initialize first
        // column with 0
        dp[i][0] = 0;
         
    for(var j = 0; j < target + 1; j++)
     
        // Initialize first
        // row with 0
        dp[0][j] = 1000000000;
         
    for(var i = 1; i <= arr.length; i++)
    {
        for(var 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]]) !=
                   1000000000 ? 
                   (dp[i][j - arr[i - 1]] + 1) : 
                   1000000000);
            }
        }
    }
 
    // Print the minimum length
    if (dp[arr.length][target] == 1000000000)
    {
        return -1;
    }
    else
    {
        return dp[arr.length][target]; 
    }
}
 
// Driver code
 
var arr = [2, 3, 5, 4, 1];
var target = 11;
var n = arr.length;
 
document.write( minlt(arr, target, n));
 
 
</script>

Output

Time Complexity: O(N*Target)
Auxiliary Space: O(N*Target)

Efficient approach: Space Optimization

In the previous approach, the dp[i][j] is depend upon the previous row and current row values but it is using space of n* target 2D matrix. So to optimize the space complexity we use a 1D vector of size target + 1 just to store the previous and current computation.

Steps that were to follow the above approach:

Create a vector ‘dp‘ of size target+1 and initialize all values to -1.
Initialize the first element of ‘dp’ to 0, since a sum of 0 can be achieved by selecting no elements.
Loop through the array ‘arr’ from the second element to the last, and for each element:
- Loop through the ‘dp’ vector from ‘target‘ down to 1.
- If the current element of ‘arr’ is greater than the current index of ‘dp’, continue to the next index.
- If the current index of ‘dp’ minus the current element of ‘arr’ is not equal to -1, update ‘dp’ with the minimum value between the current value of ‘dp’ at the
current index and ‘dp‘ at the current index minus the current element of ‘arr’ plus 1.
At last return the final answer stored in dp[target].

Below is the code to implement the above steps:

C++

// C++ code above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the smallest subarray
// with sum greater than or equal target
int minlt(vector<int> arr, int target, int n)
{
 
    // DP table to store the
    // computed subproblems
    vector<int> dp(target + 1, -1);
 
    // Initialize first column with 0
    dp[0] = 0;
 
    for (int i = 1; i <= n; i++) {
        for (int j = target; j >= 1; j--) {
 
            // Check for invalid condition
            if (arr[i - 1] > j) {
                continue;
            }
            else {
                // Fill up the dp table
                if (dp[j - arr[i - 1]] != -1) {
                    if (dp[j] == -1) {
                        dp[j] = dp[j - arr[i - 1]] + 1;
                    }
                    else {
                        dp[j] = min(dp[j],
                                    dp[j - arr[i - 1]] + 1);
                    }
                }
            }
        }
    }
 
    // Print the minimum length
    return dp[target];
}
 
// Driver code
int main()
{
    vector<int> arr = { 2, 3, 5, 4, 1 };
    int target = 11;
    int n = arr.size();
 
    cout << minlt(arr, target, n);
 
    return 0;
}

Java

import java.util.*;
 
public class Main {
    // Function to find the smallest subarray
    // with sum greater than or equal target
    public static int minlt(List<Integer> arr, int target, int n) {
 
        // DP table to store the
        // computed subproblems
        int[] dp = new int[target + 1];
        Arrays.fill(dp, -1);
 
        // Initialize first column with 0
        dp[0] = 0;
 
        for (int i = 1; i <= n; i++) {
            for (int j = target; j >= 1; j--) {
 
                // Check for invalid condition
                if (arr.get(i - 1) > j) {
                    continue;
                } else {
                    // Fill up the dp table
                    if (dp[j - arr.get(i - 1)] != -1) {
                        if (dp[j] == -1) {
                            dp[j] = dp[j - arr.get(i - 1)] + 1;
                        } else {
                            dp[j] = Math.min(dp[j], dp[j - arr.get(i - 1)] + 1);
                        }
                    }
                }
            }
        }
 
        // Print the minimum length
        return dp[target];
    }
 
    // Driver code
    public static void main(String[] args) {
        List<Integer> arr = Arrays.asList(2, 3, 5, 4, 1);
        int target = 11;
        int n = arr.size();
 
        System.out.println(minlt(arr, target, n));
    }
}

Python3

def min_subarray_with_sum(arr, target_sum):
    n = len(arr)
     
    # DP table to store the computed subproblems
    dp = [-1] * (target_sum + 1)
 
    # Initialize first column with 0
    dp[0] = 0
 
    for i in range(1, n + 1):
        for j in range(target_sum, 0, -1):
 
            # Check for invalid condition
            if arr[i - 1] > j:
                continue
            else:
                # Fill up the dp table
                if dp[j - arr[i - 1]] != -1:
                    if dp[j] == -1:
                        dp[j] = dp[j - arr[i - 1]] + 1
                    else:
                        dp[j] = min(dp[j], dp[j - arr[i - 1]] + 1)
 
    # Return the minimum length
    return dp[target_sum]
 
# Example usage:
arr = [2, 3, 5, 4, 1]
target_sum = 11
print(min_subarray_with_sum(arr, target_sum))

C#

using System;
 
namespace ConsoleApp1
{
    class Program
    {
        // 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[target + 1];
            Array.Fill(dp, -1);
 
            // Initialize first column with 0
            dp[0] = 0;
 
            for (int i = 1; i <= n; i++)
            {
                for (int j = target; j >= 1; j--)
                {
                    // Check for invalid condition
                    if (arr[i - 1] > j)
                    {
                        continue;
                    }
                    else
                    {
                        // Fill up the dp table
                        if (dp[j - arr[i - 1]] != -1)
                        {
                            if (dp[j] == -1)
                            {
                                dp[j] = dp[j - arr[i - 1]] + 1;
                            }
                            else
                            {
                                dp[j] = Math.Min(dp[j], dp[j - arr[i - 1]] + 1);
                            }
                        }
                    }
                }
            }
 
            // Print the minimum length
            return dp[target];
        }
 
        // Driver code
        static void Main(string[] args)
        {
            int[] arr = { 2, 3, 5, 4, 1 };
            int target = 11;
            int n = arr.Length;
 
            Console.WriteLine(minlt(arr, target, n));
        }
    }
}

Javascript

// Javascript code above approach
 
// Function to find the smallest subarray
// with sum greater than or equal target
function minlt(arr, target, n) {
 
    // DP table to store the
    // computed subproblems
    let dp = new Array(target + 1).fill(-1);
     
    // Initialize first column with 0
    dp[0] = 0;
     
    for (let i = 1; i <= n; i++) {
        for (let j = target; j >= 1; j--) {
     
            // Check for invalid condition
            if (arr[i - 1] > j) {
                continue;
            }
            else {
                // Fill up the dp table
                if (dp[j - arr[i - 1]] != -1) {
                    if (dp[j] == -1) {
                        dp[j] = dp[j - arr[i - 1]] + 1;
                    }
                    else {
                        dp[j] = Math.min(dp[j], dp[j - arr[i - 1]] + 1);
                    }
                }
            }
        }
    }
     
    // Print the minimum length
    return dp[target];
}
 
// Driver code
let arr = [ 2, 3, 5, 4, 1 ];
let target = 11;
let n = arr.length;
 
console.log(minlt(arr, target, n));

Output

Time Complexity: O(N*target)
Auxiliary Space: O(N*Target)

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

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