Given an array arr[] containing positive integers of size N and an integer K, the task is to make all the elements in the array equal to some value D (D > 0) such that |arr[i] – D| ? K. Print the value of D or -1 if it is not possible to make the array equal.
Examples:
Input: arr[] = {1, 2, 3, 4, 5}, K = 2
Output: 3
Explanation:
All the elements in the array can be made equal to 3 as:
|arr[0] – 3| = 2 ? 2
|arr[1] – 3| = 1 ? 2
|arr[2] – 3| = 0 ? 2
|arr[3] – 3| = 1 ? 2
|arr[4] – 3| = 2 ? 2
Input: arr[] = {1, 6, 20}, K = 1
Output: -1
Explanation:
It is not possible to equalize the array
Approach: The idea is to first find the minimum element of the array.
- If M is the minimum element of the array, then M + K is the maximum possible value which this minimum element can take in order to satisfy the condition arr[i] – D ? K.
- Then, D = M + K, if for all the elements in the array, (M + K) lies in the range [arr[i] – K, arr[i] + K].
- Therefore, we simply iterate over every element in the array and check for this condition.
Below is the implementation of the above approach:
// C++ program to make all elements // of an Array equal by adding or // subtracting at most K #include <bits/stdc++.h> using namespace std;
// Function to equalize the array by // adding or subtracting at most K // value from each element int equalize( int arr[], int n, int k)
{ // Finding the minimum element
// from the array
int min_ele
= *min_element(arr, arr + n);
// Boolean variable to check if the
// array can be equalized or not
bool flag = true ;
// Traversing the array
for ( int i = 0; i < n; i++) {
// Checking if the values lie
// within the possible range
// for each element
if (!((arr[i] + k) >= min_ele + k
&& min_ele + k >= (arr[i] - k))) {
// If any value doesn't lie in
// the range then exit the loop
flag = false ;
break ;
}
}
if (flag) {
// Value after equalizing the array
return min_ele + k;
}
// Array cannot be equalized
else
return -1;
} // Driver code int main()
{ int K = 2;
int arr[] = { 1, 2, 3, 4, 5 };
int N = sizeof (arr) / sizeof (arr[0]);
cout << equalize(arr, N, K);
} |
// Java program to make all elements // of an Array equal by adding or // subtracting at most K import java.util.*;
class GFG{
// Function to equalize the array by // adding or subtracting at most K // value from each element static int equalize( int arr[], int n, int k)
{ // Finding the minimum element
// from the array
int min_ele
= Arrays.stream(arr).min().getAsInt();
// Boolean variable to check if the
// array can be equalized or not
boolean flag = true ;
// Traversing the array
for ( int i = 0 ; i < n; i++) {
// Checking if the values lie
// within the possible range
// for each element
if (!((arr[i] + k) >= min_ele + k
&& min_ele + k >= (arr[i] - k))) {
// If any value doesn't lie in
// the range then exit the loop
flag = false ;
break ;
}
}
if (flag) {
// Value after equalizing the array
return min_ele + k;
}
// Array cannot be equalized
else
return - 1 ;
} // Driver code public static void main(String[] args)
{ int K = 2 ;
int arr[] = { 1 , 2 , 3 , 4 , 5 };
int N = arr.length;
System.out.print(equalize(arr, N, K));
} } // This code is contributed by Princi Singh |
# Python program to make all elements # of an Array equal by adding or # subtracting at most K # Function to equalize the array by # adding or subtracting at most K # value from each element def equalize(arr, n, k):
# Finding the minimum element
# from the array
min_ele = min (arr);
# Boolean variable to check if the
# array can be equalized or not
flag = True ;
# Traversing the array
for i in range (n):
# Checking if the values lie
# within the possible range
# for each element
if ( not ((arr[i] + k) > = (min_ele + k) and
(min_ele + k) > = (arr[i] - k))) :
# If any value doesn't lie in
# the range then exit the loop
flag = False ;
break ;
if (flag):
# Value after equalizing the array
return min_ele + k;
# Array cannot be equalized
else :
return - 1 ;
# Driver code if __name__ = = '__main__' :
K = 2 ;
arr = [ 1 , 2 , 3 , 4 , 5 ] ;
N = len (arr)
print (equalize(arr, N, K))
# This code is contributed by Princi Singh |
// C# program to make all elements // of an Array equal by adding or // subtracting at most K using System;
using System.Linq;
class GFG{
// Function to equalize the array by // adding or subtracting at most K // value from each element static int equalize( int []arr, int n, int k)
{ // Finding the minimum element
// from the array
int min_ele = arr.Min();
// Boolean variable to check if the
// array can be equalized or not
bool flag = true ;
// Traversing the array
for ( int i = 0; i < n; i++) {
// Checking if the values lie
// within the possible range
// for each element
if (!((arr[i] + k) >= min_ele + k
&& min_ele + k >= (arr[i] - k))) {
// If any value doesn't lie in
// the range then exit the loop
flag = false ;
break ;
}
}
if (flag) {
// Value after equalizing the array
return min_ele + k;
}
// Array cannot be equalized
else
return -1;
} // Driver code public static void Main(String[] args)
{ int K = 2;
int []arr = { 1, 2, 3, 4, 5 };
int N = arr.Length;
Console.Write(equalize(arr, N, K));
} } // This code is contributed by Princi Singh |
<script> // Javascript program to make all elements // of an Array equal by adding or // subtracting at most K // Function to equalize the array by // adding or subtracting at most K // value from each element function equalize(arr, n, k)
{ // Finding the minimum element
// from the array
let min_ele
= arr.sort((a, b) => a - b)[0];
// Boolean variable to check if the
// array can be equalized or not
let flag = true ;
// Traversing the array
for (let i = 0; i < n; i++) {
// Checking if the values lie
// within the possible range
// for each element
if (!((arr[i] + k) >= min_ele + k
&& min_ele + k >= (arr[i] - k))) {
// If any value doesn't lie in
// the range then exit the loop
flag = false ;
break ;
}
}
if (flag) {
// Value after equalizing the array
return min_ele + k;
}
// Array cannot be equalized
else
return -1;
} // Driver code let K = 2; let arr = [ 1, 2, 3, 4, 5 ]; let N = arr.length; document.write(equalize(arr, N, K)); // This code is contributed by _saurabh_jaiswal </script> |
3
Time Complexity: O(n), to traverse the array where n is the size of the given array
Auxiliary Space: O(1), as no extra space is used