Smallest subarray from a given Array with sum greater than or equal to K | Set 2
Last Updated :
25 Apr, 2023
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 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 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# 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
<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>
Time Complexity: O(2N)
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 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 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# 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
<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>
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;
}
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));
}
}
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))
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 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
3
Time Complexity: O(N*target)
Auxiliary Space: O(N*Target)
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