Minimum value of X to make all array elements equal by either decreasing or increasing by X

Given an array of N elements and you can perform two operations on it:

  • Increase any of the array element by X once.
  • Decrease any of the array element by X once.

The task is to find the minimum most value of X such that all array elements are equal by either applying operation 1, 2 or not applying any operation. If all the array elements cannot be made equal, then print -1.

Examples:

Input: a[] = {1, 4, 4, 7, 4, 1}
Output: 3
Increase the first and last element by 3.
Decrease the fourth element by 3.

Input: {1, 5, 7, 9, 1}
Output: -1



Approach: Since both the operations can only be applied once on an array element, the common point of observation is that if there are more than 3 unique elements than the answer is -1. There can arise three cases which are solved in the following ways:

  • If there are 3 unique elements, if abs(el2-el1) == abs(el3-el2), then the answer is abs(el2-el1). If they are not equal then the answer is -1.
  • If there are 2 unique elements, the answer is (el2 – el1) / 2, if el2 – el1 is even, else the answer is (el2 – el1)
  • If there is a single unique element, then the answer is 0

Below is the implementation of the above approach:

C++

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// C++ program to implement
// the above approach
  
#include <bits/stdc++.h>
using namespace std;
  
// Function that returns the
// minimum value of X
int findMinimumX(int a[], int n)
{
  
    // Declare a set
    set<int> st;
  
    // Iterate in the array element
    // and insert them into the set
    for (int i = 0; i < n; i++)
        st.insert(a[i]);
  
    // If unique elements is 1
    if (st.size() == 1)
        return 0;
  
    // Unique elements is 2
    if (st.size() == 2) {
        // Get both el2 and el1
        int el1 = *st.begin();
        int el2 = *st.rbegin();
  
        // Check if they are even
        if ((el2 - el1) % 2 == 0)
            return (el2 - el1) / 2;
        else
            return (el2 - el1);
    }
  
    // If there are 3 unique elements
    if (st.size() == 3) {
        // Get three unique elements
        auto it = st.begin();
        int el1 = *it;
        it++;
        int el2 = *it;
        it++;
        int el3 = *it;
  
        // Check if their difference is same
        if ((el2 - el1) == (el3 - el2))
            return el2 - el1;
        else
            return -1;
    }
  
    // More than 3 unique elements
    return -1;
}
  
// Driver code
int main()
{
    int a[] = { 1, 4, 4, 7, 4, 1 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << findMinimumX(a, n);
  
    return 0;
}

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Python3

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# Python 3 program to implement
# the above approach
  
# Function that returns the
# minimum value of X
def findMinimumX(a, n):
      
    # Declare a set
    st = set()
  
    # Iterate in the array element
    # and insert them into the set
    for i in range(n):
        st.add(a[i])
  
    # If unique elements is 1
    if (len(st) == 1):
        return 0
  
    # Unique elements is 2
    if (len(st) == 2):
          
        # Get both el2 and el1
        st = list(st)
        el1 = st[0]
        el2 = st[1]
  
        # Check if they are even
        if ((el2 - el1) % 2 == 0):
            return int((el2 - el1) / 2)
        else:
            return (el2 - el1)
  
    # If there are 3 unique elements
    if (len(st) == 3):
        st = list(st)
          
        # Get three unique elements
        el1 = st[0]
        el2 = st[1]
        el3 = st[2]
  
        # Check if their difference is same
        if ((el2 - el1) == (el3 - el2)):
            return el2 - el1
        else:
            return -1
  
    # More than 3 unique elements
    return -1
  
# Driver code
if __name__ == '__main__':
    a = [1, 4, 4, 7, 4, 1]
    n = len(a)
    print(findMinimumX(a, n))
  
# This code is contributed by
# Surendra_Gangwar

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Output:

3


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Striver(underscore)79 at Codechef and codeforces D

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Improved By : SURENDRA_GANGWAR