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

Given an array arr[] of integers and a number . You can change any element of the array to any 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 .

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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

an = a0 + (n-1)*d

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

We are given the values of 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 first term in the AP is 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 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 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             freq = new HashMap();            // 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 freq = new Dictionary();            // 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

 \$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 ?>

Output:

2

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