Minimum removal of elements from end of an array required to obtain sum K
Last Updated :
23 Jul, 2025
Given an integer K and an array A[] of size N, the task is to create a new array with sum K with minimum number of operations, where in each operation, an element can be removed either from the start or end of A[] and appended to the new array. If it is not possible to generate a new array with sum K, print -1. If there are multiple answers, print any one of them.
Examples
Input: K = 6, A[] = {1, 2, 3, 1, 3}, N = 5
Output: 1 3 2
Explanation: Operation 1: Removing A[0] modifies A[] to {2, 3, 1, 3}. Sum = 1.
Operation 2: Removing A[3] modifies A[] to {2, 1, 3}. Sum = 4.
Operation 3: Removing A[0] modifies A[] to {1, 3}. Sum = 6.
Input: K = 5, A[] = {1, 2, 7}, N = 3
Output: -1
Naive Approach: Follow the steps below to solve the problem:
- The task is to find two minimum length subarrays, one from the beginning and one from the end of the array (possibly empty), such that their sum is equal to K.
- Traverse the array from the left and calculate the subarray needed to be removed from the right such that the total sum is K.
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 minimum number of
// elements required to be removed from
// the ends of an array to obtain a sum K
int minSizeArr(int A[], int N, int K)
{
// Number of elements removed from the
// left and right ends of the array
int leftTaken = N, rightTaken = N;
// Sum of left and right subarrays
int leftSum = 0, rightSum = 0;
// No element is taken from left initially
for (int left = -1; left < N; left++) {
if (left != -1)
leftSum += A[left];
rightSum = 0;
// Start taking elements from right side
for (int right = N - 1; right > left; right--) {
rightSum += A[right];
if (leftSum + rightSum == K) {
// (left + 1): Count of elements
// removed from the left
// (N-right): Count of elements
// removed from the right
if (leftTaken + rightTaken
> (left + 1) + (N - right)) {
leftTaken = left + 1;
rightTaken = N - right;
}
break;
}
// If sum is greater than K
if (leftSum + rightSum > K)
break;
}
}
if (leftTaken + rightTaken <= N) {
for (int i = 0; i < leftTaken; i++)
cout << A[i] << " ";
for (int i = 0; i < rightTaken; i++)
cout << A[N - i - 1] << " ";
}
// If it is not possible to obtain sum K
else
cout << -1;
}
// Driver Code
int main()
{
int N = 7;
// Given Array
int A[] = { 3, 2, 1, 1, 1, 1, 3 };
// Given target sum
int K = 10;
minSizeArr(A, N, K);
return 0;
}
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find the minimum number of
// elements required to be removed from
// the ends of an array to obtain a sum K
static void minSizeArr(int A[], int N, int K)
{
// Number of elements removed from the
// left and right ends of the array
int leftTaken = N, rightTaken = N;
// Sum of left and right subarrays
int leftSum = 0, rightSum = 0;
// No element is taken from left initially
for (int left = -1; left < N; left++) {
if (left != -1)
leftSum += A[left];
rightSum = 0;
// Start taking elements from right side
for (int right = N - 1; right > left; right--)
{
rightSum += A[right];
if (leftSum + rightSum == K) {
// (left + 1): Count of elements
// removed from the left
// (N-right): Count of elements
// removed from the right
if (leftTaken + rightTaken
> (left + 1) + (N - right)) {
leftTaken = left + 1;
rightTaken = N - right;
}
break;
}
// If sum is greater than K
if (leftSum + rightSum > K)
break;
}
}
if (leftTaken + rightTaken <= N) {
for (int i = 0; i < leftTaken; i++)
System.out.print( A[i] + " ");
for (int i = 0; i < rightTaken; i++)
System.out.print(A[N - i - 1] + " ");
}
// If it is not possible to obtain sum K
else
System.out.print(-1);
}
// Driver code
public static void main(String[] args)
{
int N = 7;
// Given Array
int A[] = { 3, 2, 1, 1, 1, 1, 3 };
// Given target sum
int K = 10;
minSizeArr(A, N, K);
}
}
// This code is contributed by splevel62.
# Python3 program for the above approach
# Function to find the minimum number of
# elements required to be removed from
# the ends of an array to obtain a sum K
def minSizeArr(A, N, K):
# Number of elements removed from the
# left and right ends of the array
leftTaken = N
rightTaken = N
# Sum of left and right subarrays
leftSum = 0
rightSum = 0
# No element is taken from left initially
for left in range(-1, N):
if (left != -1):
leftSum += A[left]
rightSum = 0
# Start taking elements from right side
for right in range(N - 1, left, -1):
rightSum += A[right]
if (leftSum + rightSum == K):
# (left + 1): Count of elements
# removed from the left
# (N-right): Count of elements
# removed from the right
if (leftTaken + rightTaken >
(left + 1) + (N - right)):
leftTaken = left + 1
rightTaken = N - right
break
# If sum is greater than K
if (leftSum + rightSum > K):
break
if (leftTaken + rightTaken <= N):
for i in range(leftTaken):
print(A[i], end = " ")
for i in range(rightTaken):
print(A[N - i - 1], end = " ")
# If it is not possible to obtain sum K
else:
print(-1)
# Driver Code
if __name__ == "__main__":
N = 7
# Given Array
A = [ 3, 2, 1, 1, 1, 1, 3 ]
# Given target sum
K = 10
minSizeArr(A, N, K)
# This code is contributed by ukasp
// C# program for the above approach
using System;
class GFG {
// Function to find the smallest
// array that can be removed from
// the ends of an array to obtain sum K
static void minSizeArr(int[] A, int N, int K)
{
// Number of elements removed from the
// left and right ends of the array
int leftTaken = N, rightTaken = N;
// Sum of left and right subarrays
int leftSum = 0, rightSum = 0;
// No element is taken from left initially
for (int left = -1; left < N; left++) {
if (left != -1)
leftSum += A[left];
rightSum = 0;
// Start taking elements from right side
for (int right = N - 1; right > left; right--)
{
rightSum += A[right];
if (leftSum + rightSum == K) {
// (left + 1): Count of elements
// removed from the left
// (N-right): Count of elements
// removed from the right
if (leftTaken + rightTaken
> (left + 1) + (N - right)) {
leftTaken = left + 1;
rightTaken = N - right;
}
break;
}
// If sum is greater than K
if (leftSum + rightSum > K)
break;
}
}
if (leftTaken + rightTaken <= N) {
for (int i = 0; i < leftTaken; i++)
Console.Write( A[i] + " ");
for (int i = 0; i < rightTaken; i++)
Console.Write(A[N - i - 1] + " ");
}
// If it is not possible to obtain sum K
else
Console.Write(-1);
}
// Driver Code
public static void Main()
{
int N = 7;
// Given Array
int[] A = { 3, 2, 1, 1, 1, 1, 3 };
// Given target sum
int K = 10;
minSizeArr(A, N, K);
}
}
// This code is contributed by code_hunt.
<script>
// JavaScript program for the above approach
// Function to find the minimum number of
// elements required to be removed from
// the ends of an array to obtain a sum K
function minSizeArr(A, N, K)
{
// Number of elements removed from the
// left and right ends of the array
let leftTaken = N, rightTaken = N;
// Sum of left and right subarrays
let leftSum = 0, rightSum = 0;
// No element is taken from left initially
for (let left = -1; left < N; left++) {
if (left != -1)
leftSum += A[left];
rightSum = 0;
// Start taking elements from right side
for (let right = N - 1; right > left; right--)
{
rightSum += A[right];
if (leftSum + rightSum == K) {
// (left + 1): Count of elements
// removed from the left
// (N-right): Count of elements
// removed from the right
if (leftTaken + rightTaken
> (left + 1) + (N - right)) {
leftTaken = left + 1;
rightTaken = N - right;
}
break;
}
// If sum is greater than K
if (leftSum + rightSum > K)
break;
}
}
if (leftTaken + rightTaken <= N) {
for (let i = 0; i < leftTaken; i++)
document.write( A[i] + " ");
for (let i = 0; i < rightTaken; i++)
document.write(A[N - i - 1] + " ");
}
// If it is not possible to obtain sum K
else
document.write(-1);
}
// Driver code
let N = 7;
// Given Array
let A = [ 3, 2, 1, 1, 1, 1, 3 ];
// Given target sum
let K = 10;
minSizeArr(A, N, K);
// This code is contributed by souraavghosh0416.
</script>
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: Follow the steps below to optimize the above approach:
- Calculate the sum of elements of the array A[] and store it in a variable, say Total.
- The problem can be seen as finding the maximum size subarray with sum (Total - K).
- The remaining elements will add up to K.
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
// array that can be removed from
// the ends of an array to obtain sum K
void minSizeArr(int A[], int N, int K)
{
int sum = 0;
// Sum of complete array
for (int i = 0; i < N; i++)
sum += A[i];
// If given number is greater
// than sum of the array
if (K > sum) {
cout << -1;
return;
}
// If number is equal to
// the sum of array
if (K == sum) {
for (int i = 0; i < N; i++) {
cout << A[i] << " ";
}
return;
}
// tar is sum of middle subarray
int tar = sum - K;
// Find the longest subarray
// with sum equal to tar
unordered_map<int, int> um;
um[0] = -1;
int left, right;
int cur = 0, maxi = -1;
for (int i = 0; i < N; i++) {
cur += A[i];
if (um.find(cur - tar) != um.end()
&& i - um[cur - tar] > maxi) {
maxi = i - um[cur - tar];
right = i;
left = um[cur - tar];
}
if (um.find(cur) == um.end())
um[cur] = i;
}
// If there is no subarray with
// sum equal to tar
if (maxi == -1)
cout << -1;
else {
for (int i = 0; i <= left; i++)
cout << A[i] << " ";
for (int i = 0; i < right; i++)
cout << A[N - i - 1] << " ";
}
}
// Driver Code
int main()
{
int N = 7;
// Given Array
int A[] = { 3, 2, 1, 1, 1, 1, 3 };
// Given target sum
int K = 10;
minSizeArr(A, N, K);
return 0;
}
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG {
// Function to find the smallest
// array that can be removed from
// the ends of an array to obtain sum K
static void minSizeArr(int A[], int N, int K)
{
int sum = 0;
// Sum of complete array
for (int i = 0; i < N; i++)
sum += A[i];
// If given number is greater
// than sum of the array
if (K > sum) {
System.out.print(-1);
return;
}
// If number is equal to
// the sum of array
if (K == sum) {
for (int i = 0; i < N; i++) {
System.out.print(A[i] + " ");
}
return;
}
// tar is sum of middle subarray
int tar = sum - K;
// Find the longest subarray
// with sum equal to tar
HashMap<Integer, Integer> um = new HashMap<Integer, Integer>();
um.put(0, -1);
int left = 0, right = 0;
int cur = 0, maxi = -1;
for (int i = 0; i < N; i++) {
cur += A[i];
if (um.containsKey(cur - tar)
&& i - um.get(cur - tar) > maxi) {
maxi = i - um.get(cur - tar);
right = i;
left = um.get(cur - tar);
}
if (!um.containsKey(cur))
um.put(cur, i);
}
// If there is no subarray with
// sum equal to tar
if (maxi == -1)
System.out.println(-1);
else {
for (int i = 0; i <= left; i++)
System.out.print(A[i] + " ");
for (int i = 0; i < right; i++)
System.out.print(A[N - i - 1] + " ");
}
}
// Driver Code
public static void main (String[] args) {
int N = 7;
// Given Array
int A[] = { 3, 2, 1, 1, 1, 1, 3 };
// Given target sum
int K = 10;
minSizeArr(A, N, K);
}
}
// This code is contributed by Dharanendra L V.
# python 3 program for the above approach
# Function to find the smallest
# array that can be removed from
# the ends of an array to obtain sum K
def minSizeArr(A, N, K):
sum = 0
# Sum of complete array
for i in range(N):
sum += A[i]
# If given number is greater
# than sum of the array
if (K > sum):
print(-1);
return
# If number is equal to
# the sum of array
if (K == sum):
for i in range(N):
print(A[i],end = " ")
return
# tar is sum of middle subarray
tar = sum - K
# Find the longest subarray
# with sum equal to tar
um = {}
um[0] = -1
left = 0
right = 0
cur = 0
maxi = -1
for i in range(N):
cur += A[i]
if((cur - tar) in um and (i - um[cur - tar]) > maxi):
maxi = i - um[cur - tar]
right = i
left = um[cur - tar]
if(cur not in um):
um[cur] = i
# If there is no subarray with
# sum equal to tar
if (maxi == -1):
print(-1)
else:
for i in range(left+1):
print(A[i], end = " ")
for i in range(right):
print(A[N - i - 1], end = " ")
# Driver Code
if __name__ == '__main__':
N = 7
# Given Array
A = [3, 2, 1, 1, 1, 1, 3]
# Given target sum
K = 10
minSizeArr(A, N, K)
# This code is contributed by SURENDRA_GANGWAR.
// C# program for
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the smallest
// array that can be removed from
// the ends of an array to obtain sum K
static void minSizeArr(int[] A, int N, int K)
{
int sum = 0;
// Sum of complete array
for(int i = 0; i < N; i++)
sum += A[i];
// If given number is greater
// than sum of the array
if (K > sum)
{
Console.WriteLine(-1);
return;
}
// If number is equal to
// the sum of array
if (K == sum)
{
for(int i = 0; i < N; i++)
{
Console.Write(A[i] + " ");
}
return;
}
// tar is sum of middle subarray
int tar = sum - K;
// Find the longest subarray
// with sum equal to tar
Dictionary<int,
int> um = new Dictionary<int,
int>();
um[0] = -1;
int left = 0, right = 0;
int cur = 0, maxi = -1;
for(int i = 0; i < N; i++)
{
cur += A[i];
if (um.ContainsKey(cur - tar) &&
i - um[cur - tar] > maxi)
{
maxi = i - um[cur - tar];
right = i;
left = um[cur - tar];
}
if (!um.ContainsKey(cur))
um[cur] = i;
}
// If there is no subarray with
// sum equal to tar
if (maxi == -1)
Console.Write(-1);
else
{
for(int i = 0; i <= left; i++)
Console.Write(A[i] + " ");
for(int i = 0; i < right; i++)
Console.Write(A[N - i - 1] + " ");
}
}
// Driver code
static public void Main()
{
int N = 7;
// Given Array
int[] A = { 3, 2, 1, 1, 1, 1, 3 };
// Given target sum
int K = 10;
minSizeArr(A, N, K);
}
}
// This code is contributed by offbeat
<script>
// JavaScript program for the above approach
// Function to find the smallest
// array that can be removed from
// the ends of an array to obtain sum K
function minSizeArr(A, N, K)
{
var sum = 0;
var i;
// Sum of complete array
for (i = 0; i < N; i++)
sum += A[i];
// If given number is greater
// than sum of the array
if (K > sum) {
cout << -1;
return;
}
// If number is equal to
// the sum of array
if (K == sum) {
for (i = 0; i < N; i++) {
document.write(A[i]+' ');
}
return;
}
// tar is sum of middle subarray
var tar = sum - K;
// Find the longest subarray
// with sum equal to tar
var um = new Map();
um[0] = -1;
var left, right;
var cur = 0, maxi = -1;
for (i = 0; i < N; i++) {
cur += A[i];
if (um.has(cur - tar)
&& i - um.get(cur - tar) > maxi) {
maxi = i - um.get(cur - tar);
right = i;
left = um.get(cur - tar);
}
if (!um.has(cur))
um.set(cur,i);
}
// If there is no subarray with
// sum equal to tar
if (maxi == -1)
cout << -1;
else {
for (i = 0; i <= left; i++)
document.write(A[i]+' ');
for (i = 0; i < right; i++)
document.write(A[N - i - 1]+ ' ');
}
}
// Driver Code
var N = 7;
// Given Array
var A = [3, 2, 1, 1, 1, 1, 3];
// Given target sum
var K = 10;
minSizeArr(A, N, K);
</script>
Time Complexity: O(N)
Auxiliary Space: O(N)
Related Topic: Subarrays, Subsequences, and Subsets in Array
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