Minimum operations required to make two elements equal in Array
Last Updated :
23 Jul, 2025
Given array A[] of size N and integer X, the task is to find the minimum number of operations to make any two elements equal in the array. In one operation choose any element A[i] and replace it with A[i] & X. where & is bitwise AND. If such operations do not exist print -1.
Examples:
Input: A[] = {1, 2, 3, 7}, X = 3
Output: 1
Explanation: Performing above operation on A[4] = 7, A[4] = A[4] & X = 7 & 3 = 3. Array A[] becomes {1, 2, 3, 3}, 3 exists on 3rd position so in one operation two elements can be made equal in array A[].
Input: A[] = {1, 2, 3}, X = 7
Output: -1
Explanation: It is not possible to make any two elements equal by performing above operations any number of times.
Naive approach: The basic way to solve the problem is as follows:
Making bitwise AND of current element and checking whether if any other element exists with same value.
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized based on the following idea:
This problem can be divided in four cases:
- Case 1: Two equal elements already exist in array A[], In that case answer will be 0.
- Case 2: It is not possible to create two elements which are equal by above operations, In that case answer will be -1.
- Case 3: It is possible to make any two elements equal by performing above operations exactly once.
- Case 4: It is possible to make any two elements equal by performing above operations exactly twice.
Follow the steps below to solve the problem:
- Creating HashMap[].
- Iterating from 1 to N and filling frequency HashMap[] if some element repeats again return 0.
- Iterating from 1 to N and for each iteration remove the current element from HashMap[] and perform an operation on the current index and check whether it exists in HashMap[] if it exists return 1 else insert the current element again in HashMap[].
- Clear the HashMap[].
- Iterating from 1 to N and performing an operation on each element along with filling HashMap[], if the frequency of any element is more than 2 then return 2.
- Finally, if the function reaches the end return -1.
Below is the implementation of the above approach:
C++
// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;
// Function to minimum operations required
// to make two elements equal in array.
int minOperations(int A[], int N, int M)
{
// Creating HashMap
map<int, int> HashMap;
for (int i = 0; i < N; i++) {
// Filling frequency HashMap
HashMap[A[i]]++;
// If two values same already exist
if (HashMap[A[i]] >= 2)
return 0;
}
for (int i = 0; i < N; i++) {
// Removing occurrence of current
// element from HashMap
HashMap[A[i]]--;
// Performing operation on
// current index
int performOperation = A[i] & M;
// Check if another element exists
// with same value
if (HashMap[performOperation])
return 1;
// Inserting back occurrence
// of element
HashMap[A[i]]++;
}
// Clearing the HashMap
HashMap.clear();
for (int i = 0; i < N; i++) {
// Performing operation on
// current index
int performOperation = A[i] & M;
// Inserting this element
// in Hashmap
HashMap[performOperation]++;
// If two elements exist such that
// both formed with operation
if (HashMap[A[i]] == 2)
return 2;
}
// If answer do not exist
return -1;
}
// Driver Code
int main()
{
// Input 1
int A[] = { 1, 2, 3, 7 };
int N = sizeof(A) / sizeof(A[0]);
int X = 3;
// Function Call
cout << minOperations(A, N, X) << endl;
// Input 2
int A1[] = { 1, 2, 3 };
int N1 = sizeof(A1) / sizeof(A1[0]);
int X1 = 7;
// Function Call
cout << minOperations(A1, N1, X1) << endl;
return 0;
}
# Python code to implement the approach
# Function to minimum operations required
# to make two elements equal in array.
def minOperations(A, N, M):
# Creating HashMap
HashMap = {}
for i in range(N):
# If two values same already exist
if A[i] in HashMap: return 0
# Filling frequency HashMap
else: HashMap[A[i]] = 1
for i in range(N):
# Removing occurrence of current
# element from HashMap
HashMap[A[i]] -= 1
# Performing operation on
# current index
performOperation = A[i] & M
# Check if another element exists
# with same value
if (HashMap[performOperation]): return 1
# Inserting back occurrence
# of element
HashMap[A[i]] += 1
# Clearing the HashMap
HashMap = {}
for i in range(N):
# Performing operation on
# current index
performOperation = A[i] & M
# Inserting this element
# in Hashmap
if performOperation in HashMap: HashMap[performOperation] += 1
else: HashMap[performOperation] = 1
# If two elements exist such that
# both formed with operation
if (HashMap[A[i]] == 2): return 2
# If answer do not exist
return -1
# Driver Code
# Input 1
A = [1, 2, 3, 7]
N = len(A)
X = 3
# Function Call
print(minOperations(A, N, X))
# Input 2
A1 = [1, 2, 3]
N1 = len(A1)
X1 = 7
# Function Call
print(minOperations(A1, N1, X1))
//C# code to implement the approach
using System;
using System.Collections.Generic;
// Function to minimum operations required
// to make two elements equal in array.
class MainClass {
public static int MinOperations(int[] A, int N, int M) {
// Creating HashMap
Dictionary<int, int> HashMap = new Dictionary<int, int>();
for (int i = 0; i < N; i++)
{
// Filling frequency HashMap
if (HashMap.ContainsKey(A[i]))
{
HashMap[A[i]]++;
} else {
HashMap.Add(A[i], 1);
}
// If two values same already exist
if (HashMap[A[i]] >= 2)
{
return 0;
}
}
for (int i = 0; i < N; i++)
{
// Removing occurrence of current element from HashMap
HashMap[A[i]]--;
// Performing operation on current index
int performOperation = A[i] & M;
// Check if another element exists with same value
if (HashMap.ContainsKey(performOperation) &&
HashMap[performOperation] > 0) {
return 1;
}
// Inserting back occurrence of element
HashMap[A[i]]++;
}
// Clearing the HashMap
HashMap.Clear();
for (int i = 0; i < N; i++)
{
// Performing operation on current index
int performOperation = A[i] & M;
// Inserting this element in Hashmap
if (HashMap.ContainsKey(performOperation))
{
HashMap[performOperation]++;
}
else
{
HashMap.Add(performOperation, 1);
}
// If two elements exist such that both formed with operation
if (HashMap[performOperation] == 2)
{
return 2;
}
}
// If answer do not exist it will return -1
return -1;
}
// Drive Code
public static void Main(string[] args)
{
// Input 1
int[] A = { 1, 2, 3, 7 };
int N = A.Length;
int X = 3;
// Test Case 1 Function Call
Console.WriteLine(MinOperations(A, N, X));
// Input 2
int[] A1 = { 1, 2, 3 };
int N1 = A1.Length;
int X1 = 7;
// Test Case 2 Function Call
Console.WriteLine(MinOperations(A1, N1, X1));
}
}
// This Code is contributed by nikhilsainiofficial546
// JavaScript code to implement the approach
function minOperations(A, N, M) {
// Creating HashMap
let HashMap = {};
for (let i = 0; i < N; i++) {
// Filling frequency HashMap
if (HashMap[A[i]] === undefined) HashMap[A[i]] = 0;
HashMap[A[i]]++;
// If two values same already exist
if (HashMap[A[i]] >= 2) return 0;
}
for (let i = 0; i < N; i++) {
// Removing occurrence of current
// element from HashMap
HashMap[A[i]]--;
// Performing operation on
// current index
let performOperation = A[i] & M;
// Check if another element exists
// with same value
if (HashMap[performOperation]) return 1;
// Inserting back occurrence
// of element
HashMap[A[i]]++;
}
// Clearing the HashMap
HashMap = {};
for (let i = 0; i < N; i++) {
// Performing operation on
// current index
let performOperation = A[i] & M;
// Inserting this element
// in Hashmap
if (HashMap[performOperation] === undefined) HashMap[performOperation] = 0;
HashMap[performOperation]++;
// If two elements exist such that
// both formed with operation
if (HashMap[A[i]] === 2) return 2;
}
// If answer do not exist
return -1;
}
// Input 1
let A = [1, 2, 3, 7];
let N = A.length;
let X = 3;
// Function Call
console.log(minOperations(A, N, X));
// Input 2
let A1 = [1, 2, 3];
let N1 = A1.length;
let X1 = 7;
// Function Call
console.log(minOperations(A1, N1, X1));
/*package whatever //do not write package name here */
import java.io.*;
import java.util.*;
class GFG {
// Function to minimum operations required
// to make two elements equal in array.
static int minOperations(int A[], int N, int M)
{
// Creating HashMap
Map<Integer, Integer> HashMap = new HashMap<Integer, Integer>();
for (int i = 0; i < N; i++)
{
// Filling frequency HashMap
if (HashMap.containsKey(A[i]))
HashMap.put(A[i], HashMap.get(A[i]) + 1);
else
HashMap.put(A[i], 1);
// If two values same already exist
if (HashMap.get(A[i]) >= 2)
return 0;
}
for (int i = 0; i < N; i++) {
// Removing occurrence of current
// element from HashMap
if (HashMap.containsKey(A[i]))
HashMap.put(A[i], HashMap.get(A[i]) - 1);
else
HashMap.put(A[i], 1);
// Performing operation on
// current index
int performOperation = A[i] & M;
// Check if another element exists
// with same value
if (HashMap.get(performOperation)>0)
return 1;
// Inserting back occurrence
// of element
if (HashMap.containsKey(A[i]))
HashMap.put(A[i], HashMap.get(A[i]) + 1);
else
HashMap.put(A[i], 1);
}
// Clearing the HashMap
HashMap.clear();
for (int i = 0; i < N; i++) {
// Performing operation on
// current index
int performOperation = A[i] & M;
// Inserting this element
// in Hashmap
if (HashMap.containsKey(A[i]))
HashMap.put(A[i], HashMap.get(A[i]) + 1);
else
HashMap.put(A[i], 1);
// If two elements exist such that
// both formed with operation
if (HashMap.get(A[i]) == 2)
return 2;
}
// If answer do not exist
return -1;
}
// Driver Code
public static void main(String args[])
{
// Input 1
int A[] = { 1, 2, 3, 7 };
int N = A.length;
int X = 3;
// Function Call
System.out.println(minOperations(A, N, X));
// Input 2
int A1[] = { 1, 2, 3 };
int N1 = A1.length;
int X1 = 7;
// Function Call
System.out.println(minOperations(A1, N1, X1));
}
}
Time Complexity: O(N*logN)
Auxiliary Space: O(N)
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