Find whether it is possible to make array elements same using one external number | Set 2
Last Updated :
16 Sep, 2021
Given an array arr[] consisting of N integers and following are the three operations that can be performed using any external number X:
- Add X to an array element once.
- Subtract X from an array element once.
- Perform no operation on the array element.
The task is to check whether there exists a number X, such that if the above operations are performed with the number X, the array elements become equal. If the number exists, print “Yes” and the value of X. Otherwise, print “No”.
Examples:
Input: arr[] = {2, 3, 3, 4, 2}
Output: Yes 1
Explanation:
Consider the value of X as 1 and increment the array element arr[0](= 2) and arr[4](= 2) by 1, and decrement arr[3](= 4) by 1 modifies the array to {3, 3, 3, 3, 3}.
Therefore, print Yes with the value X as 1.
Input: arr[] = {4, 3, 2, 1}
Output: No
Approach: The given problem can be solved based on the following observations:
- If all numbers are equal, the answer is “YES“.
- If there are two distinct numbers in the array, the answer is “YES“, as every distinct number can be converted to another integer following the operation.
- If there are at least four distinct numbers in the array, the answer is “NO“, because of the property of the addition.
- In other cases, if there are three distinct numbers A < B < C, then all array elements can be made equal by incrementing all the As by B – A and decrementing all the Cs by C – A. Therefore, the answer will be “YES” only if 2 * B equals (C + A) / 2.
Follow the steps below to solve the problem:
- Initialize 3 variables, say, X, Y, and Z as -1 to store all the 3 distinct integers of the array arr[].
- Traverse the given array arr[] using the variable i and perform the following steps:
- If any of X, Y, and Z are -1 then assign arr[i] to that variable.
- Otherwise, if none of the X, Y, and Z are equal to arr[i], then print “NO” and return.
- If any one of X, Y, and Z is equal to -1, then print “YES“.
- Store the smallest element in X, the next largest element in Y, and the largest element in Z.
- Now, if Z-Y is equal to (Y – X), then print “YES“. Otherwise, print “NO“.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void isPossMakeThemEqual(int arr[],
int N)
{
int X = -1;
int Y = -1;
int Z = -1;
for (int i = 0; i < N; i++) {
if (X == -1) {
X = arr[i];
}
else if (X != arr[i]) {
if (Y == -1) {
Y = arr[i];
}
else if (Y != arr[i]) {
if (Z == -1) {
Z = arr[i];
}
else if (Z != arr[i]) {
cout << "NO";
return;
}
}
}
}
if (Y == -1) {
cout << "YES 0";
return;
}
if (Z == -1) {
cout << "YES " << abs(X - Y);
return;
}
int a = X, b = Y, c = Z;
X = min(a, min(b, c));
Z = max(a, max(b, c));
Y = a + b + c - X - Z;
if (Y - X != Z - Y) {
cout << "NO";
return;
}
cout << "Yes " << (Y - X);
}
int main()
{
int arr[] = { 2, 3, 3, 4, 2 };
int N = sizeof(arr) / sizeof(arr[0]);
isPossMakeThemEqual(arr, N);
return 0;
}
|
Java
import java.io.*;
class GFG {
static void isPossMakeThemEqual(int arr[],
int N)
{
int X = -1;
int Y = -1;
int Z = -1;
for (int i = 0; i < N; i++) {
if (X == -1) {
X = arr[i];
}
else if (X != arr[i]) {
if (Y == -1)
{
Y = arr[i];
}
else if (Y != arr[i]) {
if (Z == -1) {
Z = arr[i];
}
else if (Z != arr[i]) {
System.out.println("NO");
return;
}
}
}
}
if (Y == -1) {
System.out.println("YES 0");
return;
}
if (Z == -1) {
System.out.println("YES "+ Math.abs(X - Y));
return;
}
int a = X, b = Y, c = Z;
X = Math.min(a, Math.min(b, c));
Z = Math.max(a, Math.max(b, c));
Y = a + b + c - X - Z;
if (Y - X != Z - Y) {
System.out.println("NO");
return;
}
System.out.println("Yes "+(Y - X));
}
public static void main (String[] args)
{
int arr[] = { 2, 3, 3, 4, 2 };
int N =arr.length;
isPossMakeThemEqual(arr, N);
}
}
|
Python3
def isPossMakeThemEqual(arr, N):
X = -1
Y = -1
Z = -1
for i in range(N):
if (X == -1):
X = arr[i]
elif (X != arr[i]):
if (Y == -1):
Y = arr[i]
elif (Y != arr[i]):
if (Z == -1):
Z = arr[i]
elif (Z != arr[i]):
print("NO")
return
if (Y == -1):
print("YES 0")
return
if (Z == -1):
print("YES ",abs(X - Y))
return
a = X
b = Y
c = Z
X = min(a, min(b, c))
Z = max(a, max(b, c))
Y = a + b + c - X - Z
if (Y - X != Z - Y):
print("NO")
return
print("Yes ",(Y - X))
if __name__ == '__main__':
arr = [2, 3, 3, 4, 2]
N = len(arr)
isPossMakeThemEqual(arr, N)
|
C#
using System;
class GFG{
static void isPossMakeThemEqual(int[] arr,
int N)
{
int X = -1;
int Y = -1;
int Z = -1;
for (int i = 0; i < N; i++) {
if (X == -1) {
X = arr[i];
}
else if (X != arr[i]) {
if (Y == -1)
{
Y = arr[i];
}
else if (Y != arr[i]) {
if (Z == -1) {
Z = arr[i];
}
else if (Z != arr[i]) {
Console.Write("NO");
return;
}
}
}
}
if (Y == -1) {
Console.Write("YES 0");
return;
}
if (Z == -1) {
Console.Write("YES "+ Math.Abs(X - Y));
return;
}
int a = X, b = Y, c = Z;
X = Math.Min(a, Math.Min(b, c));
Z = Math.Max(a, Math.Max(b, c));
Y = a + b + c - X - Z;
if (Y - X != Z - Y) {
Console.Write("NO");
return;
}
Console.Write("Yes "+(Y - X));
}
public static void Main(String[] args)
{
int[] arr = { 2, 3, 3, 4, 2 };
int N =arr.Length;
isPossMakeThemEqual(arr, N);
}
}
|
Javascript
<script>
function isPossMakeThemEqual(arr, N)
{
let X = -1;
let Y = -1;
let Z = -1;
for (let i = 0; i < N; i++) {
if (X == -1) {
X = arr[i];
}
else if (X != arr[i]) {
if (Y == -1)
{
Y = arr[i];
}
else if (Y != arr[i]) {
if (Z == -1) {
Z = arr[i];
}
else if (Z != arr[i]) {
document.write("NO");
return;
}
}
}
}
if (Y == -1) {
document.write("YES 0");
return;
}
if (Z == -1) {
document.write("YES "+ Math.abs(X - Y));
return;
}
let a = X, b = Y, c = Z;
X = Math.min(a, Math.min(b, c));
Z = Math.max(a, Math.max(b, c));
Y = a + b + c - X - Z;
if (Y - X != Z - Y) {
document.write("NO");
return;
}
document.write("Yes "+(Y - X));
}
let arr = [ 2, 3, 3, 4, 2 ];
let N =arr.length;
isPossMakeThemEqual(arr, N);
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(1)
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