Minimum removals required such that sum of remaining array modulo M is X
Last Updated :
23 Jul, 2025
Given an array arr[] consisting of N positive integers and the integers X and M, where 0 <= X < M, the task is to find the minimum number of elements required to be removed such that sum of the remaining array modulo M is equal to X. Print -1 if not possible.
Examples:
Input: arr[] = {3, 2, 1, 2}, M = 4, X = 2
Output: 1
Explanation: One of the elements at indices (0-based) 1 or 3 can be removed. If arr[1] is removed, then arr[] modifies to {3, 1, 2} and sum % M = 6 % 4 = 2 which is equal to X = 2.
Input: arr[] = {3, 2, 1, 3}, M = 4, X = 3
Output: 1
Explanation: Remove element arr[1]( = 2). Therefore, arr[] modifies to {3, 1, 3} and sum % M = 7 % 4 = 3 which is equal to X = 3.
Naive Approach: The simplest approach is to generate all possible subsets of the given array and for each subset, check if sum modulo M of the array after removal of the subset is equal to X or not. If found to be true, store its size. Print minimum size among all such subsets obtained.
Time Complexity: O(2N) where N is the length of the given array.
Auxiliary Space: O(N)
Efficient Approach: To optimize the above approach, the idea is to use dynamic programming based on the following observations:
- If S % M > X, then the minimum number of elements having sum S % M - X must be removed from the array to make the sum modulo M equal to X.
- Otherwise, the minimum number of elements having sum S % M + M - X must be removed to make the sum modulo M equal to X.
Follow the steps below to solve the problem:
- Initialize a dp[] table, table[N + 1][X + 1] where table[i][j] represents the minimum number of elements to remove having indices in the range [0, i] such that their sum is j where X is the sum so be removed.
- Initialize dp[0][i] for each i in the range [1, X] with some large value.
- The dp transitions are as follows:
dp[i][j] = min(dp[i-1][j], dp[i][j-arr[i-1]]+1)
where, i is in the range [1, N] and j is in the range [1, X].
- Print dp[N][X] as the minimum elements to be removed.
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
// elements having sum x
int findSum(vector<int> S, int n, int x)
{
// Initialize dp table
vector<vector<int> > table(n + 1,
vector<int>(
x + 1, 0));
for (int i = 1; i <= x; i++) {
table[0][i] = INT_MAX - 1;
}
// Pre-compute subproblems
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= x; j++) {
// If mod is smaller than element
if (S[i - 1] > j) {
table[i][j] = table[i - 1][j];
}
else {
// Minimum elements with sum
// j upto index i
table[i][j]
= min(table[i - 1][j],
table[i][j
- S[i - 1]]
+ 1);
}
}
}
// Return answer
return (table[n][x] > n)
? -1
: table[n][x];
}
// Function to find minimum number
// of removals to make sum % M in
// remaining array is equal to X
void minRemovals(vector<int> arr,
int n, int m, int x)
{
// Sum of all elements
int sum = 0;
for (int i = 0; i < n; i++) {
sum += arr[i];
}
// Sum to be removed
int requied_Sum = 0;
if (sum % m < x)
requied_Sum
= m + sum % m - x;
else
requied_Sum
= sum % m - x;
// Print answer
cout << findSum(arr, n,
requied_Sum);
}
// Driver Code
int main()
{
// Given array
vector<int> arr = { 3, 2, 1, 2 };
// Given size
int n = arr.size();
// Given mod and x
int m = 4, x = 2;
// Function call
minRemovals(arr, n, m, x % m);
return 0;
}
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find the minimum
// elements having sum x
static int findSum(int[] S, int n, int x)
{
// Initialize dp table
int [][]table = new int[n + 1][x + 1];
for(int i = 1; i <= x; i++)
{
table[0][i] = Integer.MAX_VALUE - 1;
}
// Pre-compute subproblems
for(int i = 1; i <= n; i++)
{
for(int j = 1; j <= x; j++)
{
// If mod is smaller than element
if (S[i - 1] > j)
{
table[i][j] = table[i - 1][j];
}
else
{
// Minimum elements with sum
// j upto index i
table[i][j] = Math.min(table[i - 1][j],
table[i][j - S[i - 1]] + 1);
}
}
}
// Return answer
return (table[n][x] > n) ? -1 : table[n][x];
}
// Function to find minimum number
// of removals to make sum % M in
// remaining array is equal to X
static void minRemovals(int[] arr, int n,
int m, int x)
{
// Sum of all elements
int sum = 0;
for(int i = 0; i < n; i++)
{
sum += arr[i];
}
// Sum to be removed
int requied_Sum = 0;
if (sum % m < x)
requied_Sum = m + sum % m - x;
else
requied_Sum = sum % m - x;
// Print answer
System.out.print(findSum(arr, n,
requied_Sum));
}
// Driver Code
public static void main(String[] args)
{
// Given array
int[] arr = { 3, 2, 1, 2 };
// Given size
int n = arr.length;
// Given mod and x
int m = 4, x = 2;
// Function call
minRemovals(arr, n, m, x % m);
}
}
// This code is contributed by Amit Katiyar
# Python3 program for the above approach
import sys
# Function to find the minimum
# elements having sum x
def findSum(S, n, x):
# Initialize dp table
table = [[0 for x in range(x + 1)]
for y in range(n + 1)]
for i in range(1, x + 1):
table[0][i] = sys.maxsize - 1
# Pre-compute subproblems
for i in range(1, n + 1):
for j in range(1, x + 1):
# If mod is smaller than element
if (S[i - 1] > j):
table[i][j] = table[i - 1][j]
else:
# Minimum elements with sum
# j upto index i
table[i][j] = min(table[i - 1][j],
table[i][j - S[i - 1]] + 1)
# Return answer
if (table[n][x] > n):
return -1
return table[n][x]
# Function to find minimum number
# of removals to make sum % M in
# remaining array is equal to X
def minRemovals(arr, n, m, x):
# Sum of all elements
sum = 0
for i in range(n):
sum += arr[i]
# Sum to be removed
requied_Sum = 0
if (sum % m < x):
requied_Sum = m + sum % m - x
else:
requied_Sum = sum % m - x
# Print answer
print(findSum(arr, n,
requied_Sum))
# Driver Code
if __name__ == "__main__":
# Given array
arr = [ 3, 2, 1, 2 ]
# Given size
n = len(arr)
# Given mod and x
m = 4
x = 2
# Function call
minRemovals(arr, n, m, x % m)
# This code is contributed by chitranayal
// C# program for the
// above approach
using System;
class GFG{
// Function to find the minimum
// elements having sum x
static int findSum(int[] S,
int n, int x)
{
// Initialize dp table
int [,]table = new int[n + 1,
x + 1];
for(int i = 1; i <= x; i++)
{
table[0, i] = int.MaxValue - 1;
}
// Pre-compute subproblems
for(int i = 1; i <= n; i++)
{
for(int j = 1; j <= x; j++)
{
// If mod is smaller than
// element
if (S[i - 1] > j)
{
table[i, j] = table[i - 1, j];
}
else
{
// Minimum elements with sum
// j upto index i
table[i, j] = Math.Min(table[i - 1, j],
table[i, j -
S[i - 1]] + 1);
}
}
}
// Return answer
return (table[n, x] > n) ? -1 :
table[n, x];
}
// Function to find minimum number
// of removals to make sum % M in
// remaining array is equal to X
static void minRemovals(int[] arr, int n,
int m, int x)
{
// Sum of all elements
int sum = 0;
for(int i = 0; i < n; i++)
{
sum += arr[i];
}
// Sum to be removed
int requied_Sum = 0;
if (sum % m < x)
requied_Sum = m + sum %
m - x;
else
requied_Sum = sum % m - x;
// Print answer
Console.Write(findSum(arr, n,
requied_Sum));
}
// Driver Code
public static void Main(String[] args)
{
// Given array
int[] arr = {3, 2, 1, 2};
// Given size
int n = arr.Length;
// Given mod and x
int m = 4, x = 2;
// Function call
minRemovals(arr, n, m, x % m);
}
}
// This code is contributed by Amit Katiyar
<script>
// JavaScript program to implement
// the above approach
// Function to find the minimum
// elements having sum x
function findSum(S, n, x)
{
// Initialize dp table
let table = new Array(n + 1);
// Loop to create 2D array using 1D array
for (var i = 0; i < table.length; i++) {
table[i] = new Array(2);
}
for (var i = 0; i < table.length; i++) {
for (var j = 0; j < table.length; j++) {
table[i][j] = 0;
}
}
for(let i = 1; i <= x; i++)
{
table[0][i] = Number.MAX_VALUE - 1;
}
// Pre-compute subproblems
for(let i = 1; i <= n; i++)
{
for(let j = 1; j <= x; j++)
{
// If mod is smaller than element
if (S[i - 1] > j)
{
table[i][j] = table[i - 1][j];
}
else
{
// Minimum elements with sum
// j upto index i
table[i][j] = Math.min(table[i - 1][j],
table[i][j - S[i - 1]] + 1);
}
}
}
// Return answer
return (table[n][x] > n) ? -1 : table[n][x];
}
// Function to find minimum number
// of removals to make sum % M in
// remaining array is equal to X
function minRemovals(arr, n, m, x)
{
// Sum of all elements
let sum = 0;
for(let i = 0; i < n; i++)
{
sum += arr[i];
}
// Sum to be removed
let requied_Sum = 0;
if (sum % m < x)
requied_Sum = m + sum % m - x;
else
requied_Sum = sum % m - x;
// Print answer
document.write(findSum(arr, n,
requied_Sum));
}
// Driver Code
// Given array
let arr = [ 3, 2, 1, 2 ];
// Given size
let n = arr.length;
// Given mod and x
let m = 4, x = 2;
// Function call
minRemovals(arr, n, m, x % m);
</script>
Time Complexity: O(N*X) where N is the length of the given array and X is the given integer.
Auxiliary Space: O(N*X)
Efficient approach : Space optimization
In previous approach we use 2d matrix to store the computation of subproblems but the current computation is only depend upon the previous row and con current row so to optimize the space complexity we use a 1D vector of size X+1 to get the computation of previous row.
Implementation steps :
- Create a vectors table of size x+1 to store previous row computation of matrix.
- For setting the Base Case initialize the table with INT_MAX - 1.
- Now iterate over subproblem and store the current with the help of table.
- Update the current value of table previous computations.
- At last check if answer exists and return table[x] else return -1.
Implementation:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the minimum
// elements having sum x
int findSum(vector<int> S, int n, int x)
{
// Initialize dp table
vector<int> table(x + 1, 0);
for (int i = 1; i <= x; i++) {
table[i] = INT_MAX - 1;
}
// Pre-compute subproblems
for (int i = 1; i <= n; i++) {
for (int j = x; j >= 1; j--) {
// If mod is smaller than element
if (S[i - 1] > j) {
continue;
}
else {
// Minimum elements with sum
// j upto index i
table[j] = min(table[j],
table[j - S[i - 1]]
+ 1);
}
}
}
// Return answer
return (table[x] > n)
? -1
: table[x];
}
// Function to find minimum number
// of removals to make sum % M in
// remaining array is equal to X
void minRemovals(vector<int> arr,
int n, int m, int x)
{
// Sum of all elements
int sum = 0;
for (int i = 0; i < n; i++) {
sum += arr[i];
}
// Sum to be removed
int requied_Sum = 0;
if (sum % m < x)
requied_Sum
= m + sum % m - x;
else
requied_Sum
= sum % m - x;
// Print answer
cout << findSum(arr, n,
requied_Sum);
}
// Driver Code
int main()
{
// Given array
vector<int> arr = { 3, 2, 1, 2 };
// Given size
int n = arr.size();
// Given mod and x
int m = 4, x = 2;
// Function call
minRemovals(arr, n, m, x % m);
return 0;
}
// this code is contributed by bhardwajji
import java.util.*;
class Main {
// Function to find the minimum
// elements having sum x
public static int findSum(List<Integer> S, int n, int x) {
// Initialize dp table
int[] table = new int[x + 1];
Arrays.fill(table, Integer.MAX_VALUE - 1);
table[0] = 0;
// Pre-compute subproblems
for (int i = 1; i <= n; i++) {
for (int j = x; j >= 1; j--) {
// If mod is smaller than element
if (S.get(i - 1) > j) {
continue;
} else {
// Minimum elements with sum
// j upto index i
table[j] = Math.min(table[j],
table[j - S.get(i - 1)]
+ 1);
}
}
}
// Return answer
return (table[x] > n) ? -1 : table[x];
}
// Function to find minimum number
// of removals to make sum % M in
// remaining array is equal to X
public static void minRemovals(List<Integer> arr,
int n, int m, int x) {
// Sum of all elements
int sum = 0;
for (int i = 0; i < n; i++) {
sum += arr.get(i);
}
// Sum to be removed
int required_Sum = 0;
if (sum % m < x) {
required_Sum = m + sum % m - x;
} else {
required_Sum = sum % m - x;
}
// Print answer
System.out.println(findSum(arr, n, required_Sum));
}
// Driver code
public static void main(String[] args) {
// Given array
List<Integer> arr = new ArrayList<>();
arr.add(3);
arr.add(2);
arr.add(1);
arr.add(2);
// Given size
int n = arr.size();
// Given mod and x
int m = 4, x = 2;
// Function call
minRemovals(arr, n, m, x % m);
}
}
# Function to find the minimum
# elements having sum x
def findSum(S, n, x):
# Initialize dp table
table = [0] * (x + 1)
for i in range(1, x+1):
table[i] = float('inf')
# Pre-compute subproblems
for i in range(1, n+1):
for j in range(x, 0, -1):
# If mod is smaller than element
if S[i - 1] > j:
continue
else:
# Minimum elements with sum
# j upto index i
table[j] = min(table[j], table[j - S[i - 1]] + 1)
# Return answer
return -1 if table[x] > n else table[x]
# Function to find minimum number
# of removals to make sum % M in
# remaining array is equal to X
def minRemovals(arr, n, m, x):
# Sum of all elements
sum_arr = sum(arr)
# Sum to be removed
required_sum = (m + sum_arr % m - x) % m if sum_arr % m < x else (sum_arr % m - x)
# Print answer
print(findSum(arr, n, required_sum))
# Driver Code
if __name__ == '__main__':
# Given array
arr = [3, 2, 1, 2]
# Given size
n = len(arr)
# Given mod and x
m = 4
x = 2
# Function call
minRemovals(arr, n, m, x % m)
using System;
class GFG {
// Function to find the minimum
// elements having sum x
static int FindSum(int[] S, int n, int x)
{
// Initialize dp table
int[] table = new int[x + 1];
for (int i = 1; i <= x; i++)
table[i] = int.MaxValue - 1;
// Pre-compute subproblems
for (int i = 1; i <= n; i++) {
for (int j = x; j >= 1; j--) {
// If mod is smaller than element
if (S[i - 1] > j)
continue;
else {
// Minimum elements with sum
// j upto index i
table[j] = Math.Min(
table[j], table[j - S[i - 1]] + 1);
}
}
}
// Return answer
return (table[x] > n) ? -1 : table[x];
}
// Function to find minimum number
// of removals to make sum % M in
// remaining array is equal to X
static void MinRemovals(int[] arr, int n, int m, int x)
{
// Sum of all elements
int sum = 0;
for (int i = 0; i < n; i++)
sum += arr[i];
// Sum to be removed
int required_sum = 0;
if (sum % m < x)
required_sum = m + sum % m - x;
else
required_sum = sum % m - x;
// Print answer
Console.WriteLine(FindSum(arr, n, required_sum));
}
// Driver Code
public static void Main()
{
// Given array
int[] arr = { 3, 2, 1, 2 };
// Given size
int n = arr.Length;
// Given mod and x
int m = 4, x = 2;
// Function call
MinRemovals(arr, n, m, x % m);
}
}
// This code is contributed by sarojmcy2e
// Function to find the minimum
// elements having sum x
function findSum(S, n, x) {
// Initialize dp table
let table = new Array(x + 1).fill(0);
for (let i = 1; i <= x; i++) {
table[i] = Number.MAX_SAFE_INTEGER - 1;
}
// Pre-compute subproblems
for (let i = 1; i <= n; i++) {
for (let j = x; j >= 1; j--) {
// If mod is smaller than element
if (S[i - 1] > j) {
continue;
} else {
// Minimum elements with sum
// j upto index i
table[j] = Math.min(table[j], table[j - S[i - 1]] + 1);
}
}
}
// Return answer
return table[x] > n ? -1 : table[x];
}
// Function to find minimum number
// of removals to make sum % M in
// remaining array is equal to X
function minRemovals(arr, n, m, x) {
// Sum of all elements
let sum = arr.reduce((a, b) => a + b, 0);
// Sum to be removed
let requiredSum = sum % m < x ? m + sum % m - x : sum % m - x;
// Print answer
console.log(findSum(arr, n, requiredSum));
}
let arr = [3, 2, 1, 2];
let n = arr.length;
let m = 4;
let x = 2;
minRemovals(arr, n, m, x % m);
Time Complexity: O(N*X)
Auxiliary Space: O(X)
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