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Minimum number of changes required to make the given array an AP

  • Difficulty Level : Medium
  • Last Updated : 14 May, 2021

Given an array arr[] of N  integers and a number d. You can change any element of the array to an integer. The task is to find the minimum number of change(s) required to make the given array an Arithmetic Progression with the common difference d.

Examples:  

Input : N = 4, d = 2
        arr[] = {1, 2, 4, 6}
Output : 1
Explanation: change a[0]=0. 
So, new sequence is 0, 2, 4, 6 which is an AP.

Input : N = 5, d = 1
        arr[] = {1, 3, 3, 4, 6}
Output : 2
Explanation: change a[1]=2 and a[4]=5. 
So, new sequence is 1, 2, 3, 4, 5 which is an AP.
 

The idea to solve this problem is to observe that the formulae for the n-th term in an AP are:  

an = a0 + (n-1)*d

Where, a0 is the first term
and d is the common difference.

We are given the values of d  and an. So, we will find the value of a0 for all values of i, where 1<=i<=n and store the frequency of occurrences of a0 for different values of i.
Now, the minimum number of elements needed to be changed is:  

n - (maximum frequency of a0)

Where maximum frequency of a0 signifies the total number of elements in the array for which the value of the first term in the AP is the same.



Below is the implementation of the above approach: 

C++




// C++ program to find the minimum number
// of changes required to make the given
// array an AP with common difference d
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum number
// of changes required to make the given
// array an AP with common difference d
int minimumChanges(int arr[], int n, int d)
{
    int maxFreq = INT_MIN;
 
    // Map to store frequency of a0
    unordered_map<int, int> freq;
 
    // storing frequency of a0 for all possible
    // values of a[i] and finding the maximum
    // frequency
    for (int i = 0; i < n; ++i) {
        int a0 = arr[i] - (i)*d;
 
        // increment frequency by 1
        if (freq.find(a0) != freq.end()) {
            freq[a0]++;
        }
        else
            freq.insert(make_pair(a0, 1));
 
        // finds count of most frequent number
        if (freq[a0] > maxFreq)
            maxFreq = freq[a0];
    }
 
    // minimum number of elements needed to
    // be changed is: n - (maximum frequency of a0)
    return (n - maxFreq);
}
 
// Driver Program
int main()
{
    int n = 5, d = 1;
 
    int arr[] = { 1, 3, 3, 4, 6 };
 
    cout << minimumChanges(arr, n, d);
 
    return 0;
}

Java




// Java program to find the
// minimum number of changes
// required to make the given
// array an AP with common
// difference d
import java.util.*;
 
class GFG {
 
    // Function to find the minimum
    // number of changes required
    // to make the given array an
    // AP with common difference d
    static int minimumChanges(int arr[],
                              int n, int d)
    {
        int maxFreq = -1;
 
        // Map to store frequency of a0
        HashMap<Integer,
                Integer>
            freq = new HashMap<Integer,
                               Integer>();
 
        // storing frequency of a0 for
        // all possible values of a[i]
        // and finding the maximum
        // frequency
        for (int i = 0; i < n; ++i) {
            int a0 = arr[i] - (i)*d;
 
            // increment frequency by 1
            if (freq.containsKey(a0)) {
                freq.put(a0, freq.get(a0) + 1);
            }
            else
                freq.put(a0, 1);
 
            // finds count of most
            // frequent number
            if (freq.get(a0) > maxFreq)
                maxFreq = freq.get(a0);
        }
 
        // minimum number of elements
        // needed to be changed is:
        // n - (maximum frequency of a0)
        return (n - maxFreq);
    }
 
    // Driver Code
    public static void main(String args[])
    {
        int n = 5, d = 1;
 
        int arr[] = { 1, 3, 3, 4, 6 };
 
        System.out.println(minimumChanges(arr, n, d));
    }
}
 
// This code is contributed
// by Arnab Kundu

Python3




# Python3 program to find the minimum
# number of changes required to make
# the given array an AP with common
# difference d
 
# Function to find the minimum number
# of changes required to make the given
# array an AP with common difference d
def minimumChanges(arr, n, d):
    maxFreq = -2147483648
     
    # dictionary to store
    # frequency of a0
    freq = {}
     
    # storing frequency of a0 for
    # all possible values of a[i]
    # and finding the maximum'
    # frequency
    for i in range(n):
        a0 = arr[i] - i * d
         
        # increment frequency by 1
        if a0 in freq:
            freq[a0] += 1
        else:
            freq[a0] = 1
             
        # finds count of most
        # frequent number
        if freq[a0] > maxFreq:
            maxFreq = freq[a0]
             
    # minimum number of elements
    # needed to be changed is:
    # n - (maximum frequency of a0)
    return (n-maxFreq)
 
# Driver Code
 
# number of terms in ap
n = 5
 
# difference in AP
d = 1
arr = [1, 3, 3, 4, 6 ]
ans = minimumChanges(arr, n, d)
print(ans)
 
# This code is contributed
# by sahil shelangia

C#




// C# program to find the
// minimum number of changes
// required to make the given
// array an AP with common
// difference d
using System;
using System.Collections.Generic;
 
class GFG {
 
    // Function to find the minimum
    // number of changes required
    // to make the given array an
    // AP with common difference d
    static int minimumChanges(int[] arr,
                              int n, int d)
    {
        int maxFreq = -1;
 
        // Map to store frequency of a0
        Dictionary<int, int> freq = new Dictionary<int, int>();
 
        // storing frequency of a0 for
        // all possible values of a[i]
        // and finding the maximum
        // frequency
        for (int i = 0; i < n; ++i) {
            int a0 = arr[i] - (i)*d;
 
            // increment frequency by 1
            if (freq.ContainsKey(a0)) {
                var obj = freq[a0];
                freq.Remove(a0);
                freq.Add(a0, obj + 1);
            }
            else
                freq.Add(a0, 1);
 
            // finds count of most
            // frequent number
            if (freq[a0] > maxFreq)
                maxFreq = freq[a0];
        }
 
        // minimum number of elements
        // needed to be changed is:
        // n - (maximum frequency of a0)
        return (n - maxFreq);
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int n = 5, d = 1;
 
        int[] arr = { 1, 3, 3, 4, 6 };
 
        Console.WriteLine(minimumChanges(arr, n, d));
    }
}
 
// This code contributed by Rajput-Ji

PHP




<?php
// PHP program to find the minimum number
// of changes required to make the given
// array an AP with common difference d
 
// Function to find the minimum number
// of changes required to make the given
// array an AP with common difference d
function minimumChanges(&$arr, $n, $d)
{
    $maxFreq = PHP_INT_MIN;
     
    // array to store frequency of a0
    $freq = array();
     
    // storing frequency of a0 for
    // all possible values of a[i]
    // and finding the maximum
    // frequency
    for ($i = 0; $i < $n; ++$i)
    {
        $a0 = $arr[$i] - ($i) * $d;
 
        // increment frequency by 1
        if(array_search($a0, $freq))
        {
            $freq[$a0]++;
        }
        else
            $freq[$a0] = 1;
 
        // finds count of most
        // frequent number
        if ($freq[$a0] > $maxFreq)
            $maxFreq = $freq[$a0];
    }
     
    // minimum number of elements
    // needed to be changed is:
    // $n - (maximum frequency of a0)
    return ($n - $maxFreq);
}
 
// Driver Code
$n = 5;
$d = 1;
     
$arr = array( 1, 3, 3, 4, 6 );
     
echo minimumChanges($arr, $n, $d);
     
// This code is contributed
// by ChitraNayal
?>

Javascript




<script>
// Javascript program to find the
// minimum number of changes
// required to make the given
// array an AP with common
// difference d
     
    // Function to find the minimum
    // number of changes required
    // to make the given array an
    // AP with common difference d
    function minimumChanges(arr,n,d)
    {
        let maxFreq = -1;
   
        // Map to store frequency of a0
        let freq = new Map();
   
        // storing frequency of a0 for
        // all possible values of a[i]
        // and finding the maximum
        // frequency
        for (let i = 0; i < n; ++i) {
            let a0 = arr[i] - (i)*d;
   
            // increment frequency by 1
            if (freq.has(a0)) {
                freq.set(a0, freq.get(a0) + 1);
            }
            else
                freq.set(a0, 1);
   
            // finds count of most
            // frequent number
            if (freq.get(a0) > maxFreq)
                maxFreq = freq.get(a0);
        }
   
        // minimum number of elements
        // needed to be changed is:
        // n - (maximum frequency of a0)
        return (n - maxFreq);
    }
     
      // Driver Code
    let n = 5, d = 1;
    let arr=[ 1, 3, 3, 4, 6];
    document.write(minimumChanges(arr, n, d));
     
 
// This code is contributed by avanitrachhadiya2155
</script>
Output: 
2

 

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