Minimize count of array elements to be removed such that at least K elements are equal to their index values
Last Updated :
23 Jul, 2025
Given an array arr[](1- based indexing) consisting of N integers and a positive integer K, the task is to find the minimum number of array elements that must be removed such that at least K array elements are equal to their index values. If it is not possible to do so, then print -1.
Examples:
Input: arr[] = {5, 1, 3, 2, 3} K = 2
Output: 2
Explanation:
Following are the removal operations required:
- Removing arr[1] modifies array to {1, 3, 2, 3} -> 1 element is equal to its index value.
- Removing arr[3] modifies array to {1, 2, 3} -> 3 elements are equal to their index value.
After the above operations 3(>= K) elements are equal to their index values and the minimum removals required is 2.
Input: arr[] = {2, 3, 4} K = 1
Output: -1
Approach: The above problem can be solved with the help of Dynamic Programming. Follow the steps below to solve the given problem.
- Initialize a 2-D dp table such that dp[i][j] will denote maximum elements that have values equal to their indexes when a total of j elements are present.
- All the values in the dp table are initially filled with 0s.
- Iterate for each i in the range [0, N-1] and j in the range [0, i], there are two choices.
- Delete the current element, the dp table can be updated as dp[i+1][j] = max(dp[i+1][j], dp[i][j]).
- Keep the current element, then dp table can be updated as: dp[i+1][j+1] = max(dp[i+1][j+1], dp[i][j] + (arr[i+1] == j+1)).
- Now for each j in the range [N, 0] check if the value of dp[N][j] is greater than or equal to K. Take minimum if found and return the answer.
- Otherwise, return -1 at the end. That means no possible answer is found.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to minimize the removals of
// array elements such that atleast K
// elements are equal to their indices
int MinimumRemovals(int a[], int N, int K)
{
// Store the array as 1-based indexing
// Copy of first array
int b[N + 1];
for (int i = 0; i < N; i++) {
b[i + 1] = a[i];
}
// Make a dp-table of (N*N) size
int dp[N + 1][N + 1];
// Fill the dp-table with zeroes
memset(dp, 0, sizeof(dp));
for (int i = 0; i < N; i++) {
for (int j = 0; j <= i; j++) {
// Delete the current element
dp[i + 1][j] = max(
dp[i + 1][j], dp[i][j]);
// Take the current element
dp[i + 1][j + 1] = max(
dp[i + 1][j + 1],
dp[i][j] + ((b[i + 1] == j + 1) ? 1 : 0));
}
}
// Check for the minimum removals
for (int j = N; j >= 0; j--) {
if (dp[N][j] >= K) {
return (N - j);
}
}
return -1;
}
// Driver Code
int main()
{
int arr[] = { 5, 1, 3, 2, 3 };
int K = 2;
int N = sizeof(arr) / sizeof(arr[0]);
cout << MinimumRemovals(arr, N, K);
return 0;
}
// Java program for the above approach
import java.io.*;
class GFG {
// Function to minimize the removals of
// array elements such that atleast K
// elements are equal to their indices
static int MinimumRemovals(int a[], int N, int K)
{
// Store the array as 1-based indexing
// Copy of first array
int b[] = new int[N + 1];
for (int i = 0; i < N; i++) {
b[i + 1] = a[i];
}
// Make a dp-table of (N*N) size
int dp[][] = new int[N + 1][N + 1];
for (int i = 0; i < N; i++) {
for (int j = 0; j <= i; j++) {
// Delete the current element
dp[i + 1][j] = Math.max(dp[i + 1][j], dp[i][j]);
// Take the current element
dp[i + 1][j + 1] = Math.max(
dp[i + 1][j + 1],
dp[i][j]
+ ((b[i + 1] == j + 1) ? 1 : 0));
}
}
// Check for the minimum removals
for (int j = N; j >= 0; j--) {
if (dp[N][j] >= K) {
return (N - j);
}
}
return -1;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 5, 1, 3, 2, 3 };
int K = 2;
int N = arr.length;
System.out.println(MinimumRemovals(arr, N, K));
}
}
// This code is contributed by Dharanendra L V.
# Python 3 program for the above approach
# Function to minimize the removals of
# array elements such that atleast K
# elements are equal to their indices
def MinimumRemovals(a, N, K):
# Store the array as 1-based indexing
# Copy of first array
b = [0 for i in range(N + 1)]
for i in range(N):
b[i + 1] = a[i]
# Make a dp-table of (N*N) size
dp = [[0 for i in range(N+1)] for j in range(N+1)]
for i in range(N):
for j in range(i + 1):
# Delete the current element
dp[i + 1][j] = max(dp[i + 1][j], dp[i][j])
# Take the current element
dp[i + 1][j + 1] = max(dp[i + 1][j + 1],dp[i][j] + (1 if (b[i + 1] == j + 1) else 0))
# Check for the minimum removals
j = N
while(j >= 0):
if(dp[N][j] >= K):
return (N - j)
j -= 1
return -1
# Driver Code
if __name__ == '__main__':
arr = [5, 1, 3, 2, 3]
K = 2
N = len(arr)
print(MinimumRemovals(arr, N, K))
# This code is contributed by SURENDRA_GANGWAR.
// C# code for the above approach
using System;
public class GFG
{
// Function to minimize the removals of
// array elements such that atleast K
// elements are equal to their indices
static int MinimumRemovals(int[] a, int N, int K)
{
// Store the array as 1-based indexing
// Copy of first array
int[] b = new int[N + 1];
for (int i = 0; i < N; i++) {
b[i + 1] = a[i];
}
// Make a dp-table of (N*N) size
int[, ] dp = new int[N + 1, N + 1];
for (int i = 0; i < N; i++) {
for (int j = 0; j <= i; j++) {
// Delete the current element
dp[i + 1, j]
= Math.Max(dp[i + 1, j], dp[i, j]);
// Take the current element
dp[i + 1, j + 1] = Math.Max(
dp[i + 1, j + 1],
dp[i, j]
+ ((b[i + 1] == j + 1) ? 1 : 0));
}
}
// Check for the minimum removals
for (int j = N; j >= 0; j--) {
if (dp[N, j] >= K) {
return (N - j);
}
}
return -1;
}
static public void Main()
{
// Code
int[] arr = { 5, 1, 3, 2, 3 };
int K = 2;
int N = arr.Length;
Console.Write(MinimumRemovals(arr, N, K));
}
}
// This code is contributed by Potta Lokesh
<script>
// Javascript program for the above approach
// Function to minimize the removals of
// array elements such that atleast K
// elements are equal to their indices
function MinimumRemovals(a, N, K)
{
// Store the array as 1-based indexing
// Copy of first array
let b = new Array(N + 1);
for (let i = 0; i < N; i++) {
b[i + 1] = a[i];
}
// Make a dp-table of (N*N) size
let dp = new Array(N + 1).fill(0).map(() => new Array(N + 1).fill(0));
for (let i = 0; i < N; i++) {
for (let j = 0; j <= i; j++) {
// Delete the current element
dp[i + 1][j] = Math.max(
dp[i + 1][j], dp[i][j]);
// Take the current element
dp[i + 1][j + 1] = Math.max(
dp[i + 1][j + 1],
dp[i][j] + ((b[i + 1] == j + 1) ? 1 : 0));
}
}
// Check for the minimum removals
for (let j = N; j >= 0; j--) {
if (dp[N][j] >= K) {
return (N - j);
}
}
return -1;
}
// Driver Code
let arr = [ 5, 1, 3, 2, 3 ];
let K = 2;
let N = arr.length
document.write(MinimumRemovals(arr, N, K));
// This code is contributed by _saurabh_jaiswal.
</script>
Time Complexity: O(N2)
Auxiliary Space: O(N2)
Efficient approach : Space optimization
In previous approach the current value dp[i][j] is only depend upon the current and previous row values of DP. So to optimize the space complexity we use a single 1D array to store the computations.
Implementation steps:
- Create a 1D vector dp of size N+1 and initialize it with 0.
- Set a base case by initializing the values of DP .
- Now iterate over subproblems by the help of nested loop and get the current value from previous computations.
- At last iterate over Dp and whenever dp[j] >= K return (N - j) .
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
// Function to minimize the removals of
// array elements such that atleast K
// elements are equal to their indices
int MinimumRemovals(int a[], int N, int K)
{
// Store the array as 1-based indexing
// Copy of first array
int b[N + 1];
for (int i = 0; i < N; i++) {
b[i + 1] = a[i];
}
// Initialize the dp-table with zeroes
int dp[N + 1];
memset(dp, 0, sizeof(dp));
// Iterate over the array and fill the dp-table
for (int i = 0; i < N; i++) {
for (int j = i + 1; j > 0; j--) {
dp[j] = max(dp[j], dp[j - 1] + ((b[i + 1] == j) ? 1 : 0));
}
}
// Check for the minimum removals
for (int j = N; j >= 0; j--) {
if (dp[j] >= K) {
return (N - j);
}
}
return -1;
}
// Driver Code
int main()
{
int arr[] = { 5, 1, 3, 2, 3 };
int K = 2;
int N = sizeof(arr) / sizeof(arr[0]);
cout << MinimumRemovals(arr, N, K);
return 0;
}
// this code is contributed by bhardwajji
import java.util.Arrays;
public class MinimumRemovals {
// Function to minimize the removals of
// array elements such that atleast K
// elements are equal to their indices
static int minimumRemovals(int a[], int N, int K)
{
// Store the array as 1-based indexing
// Copy of first array
int b[] = new int[N + 1];
for (int i = 0; i < N; i++) {
b[i + 1] = a[i];
}
// Initialize the dp-table with zeroes
int dp[] = new int[N + 1];
Arrays.fill(dp, 0);
// Iterate over the array and fill the dp-table
for (int i = 0; i < N; i++) {
for (int j = i + 1; j > 0; j--) {
dp[j] = Math.max(dp[j], dp[j - 1] + ((b[i + 1] == j) ? 1 : 0));
}
}
// Check for the minimum removals
for (int j = N; j >= 0; j--) {
if (dp[j] >= K) {
return (N - j);
}
}
return -1;
}
// Driver Code
public static void main(String[] args) {
int arr[] = { 5, 1, 3, 2, 3 };
int K = 2;
int N = arr.length;
System.out.println(minimumRemovals(arr, N, K));
}
}
# Function to minimize the removals of
# array elements such that atleast K
# elements are equal to their indices
def MinimumRemovals(a, N, K):
# Store the array as 1-based indexing
# Copy of first array
b = [0] * (N + 1)
for i in range(N):
b[i + 1] = a[i]
# Initialize the dp-table with zeroes
dp = [0] * (N + 1)
# Iterate over the array and fill the dp-table
for i in range(N):
for j in range(i + 1, 0, -1):
dp[j] = max(dp[j], dp[j - 1] + ((b[i + 1] == j)))
# Check for the minimum removals
for j in range(N, -1, -1):
if dp[j] >= K:
return (N - j)
return -1
# Driver Code
arr = [5, 1, 3, 2, 3]
K = 2
N = len(arr)
print(MinimumRemovals(arr, N, K))
using System;
public class Program
{
// Function to minimize the removals of
// array elements such that atleast K
// elements are equal to their indices
public static int MinimumRemovals(int[] a, int N, int K)
{
// Store the array as 1-based indexing
// Copy of first array
int[] b = new int[N + 1];
for (int i = 0; i < N; i++) {
b[i + 1] = a[i];
}
// Initialize the dp-table with zeroes
int[] dp = new int[N + 1];
Array.Fill(dp, 0);
// Iterate over the array and fill the dp-table
for (int i = 0; i < N; i++) {
for (int j = i + 1; j > 0; j--) {
dp[j] = Math.Max(
dp[j],
dp[j - 1] + ((b[i + 1] == j) ? 1 : 0));
}
}
// Check for the minimum removals
for (int j = N; j >= 0; j--) {
if (dp[j] >= K) {
return (N - j);
}
}
return -1;
}
// Driver Code
public static void Main()
{
int[] arr = { 5, 1, 3, 2, 3 };
int K = 2;
int N = arr.Length;
Console.WriteLine(MinimumRemovals(arr, N, K));
}
}
// This code is contributed by user_dtewbxkn77n
function minimumRemovals(a, N, K) {
// Store the array as 1-based indexing
// Copy of the first array
let b = new Array(N + 1);
for (let i = 0; i < N; i++) {
b[i + 1] = a[i];
}
// Initialize the dp-table with zeroes
let dp = new Array(N + 1).fill(0);
// Iterate over the array and fill the dp-table
for (let i = 0; i < N; i++) {
for (let j = i + 1; j > 0; j--) {
dp[j] = Math.max(dp[j], dp[j - 1] + (b[i + 1] === j ? 1 : 0));
}
}
// Check for the minimum removals
for (let j = N; j >= 0; j--) {
if (dp[j] >= K) {
return N - j;
}
}
return -1;
}
// Driver Code
let arr = [5, 1, 3, 2, 3];
let K = 2;
let N = arr.length;
console.log(minimumRemovals(arr, N, K));
Output:
2
Time Complexity: O(N^2)
Auxiliary Space: O(N)
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