Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Minimum operations required to transform a sequence of numbers to a sequence where a[i]=a[i+2]

  • Difficulty Level : Basic
  • Last Updated : 25 May, 2021

Given a sequence of integers of even length ‘n’, the task is to find the minimum number of operations required to convert the sequence to follow the rule a[i]=a[i+2] where ‘i’ is the index. 
The operation here is to replace any element of the sequence with any element.

Examples:  

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Input : n=4 ; Array : 3, 1, 3, 2
Output : 1
If we change the last element to '1' then, 
the sequence will become 3, 1, 3, 1 (satisfying the condition)
So, only 1 replacement is required.

Input : n=6 ; Array : 105 119 105 119 105 119
Output : 0
As the sequence is already in the required state.
So, no replacement of elements is required. 

Approach: As we see that the indices 0, 2, …, n-2 are connected independently and 1, 3, 5, …, n are connected independently and must have the same value. So, 

  • We have to find the most occurring number in both the sequences (even and odd) by storing the numbers and their frequency in a map.
  • Then every other number in that sequence will have to be replaced with the most occurring number in the same sequence.
  • Finally, the count of replacements from the previous step will be the answer.

Below is the implementation of the above approach:  

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the minimum replacements
int minReplace(int a[], int n)
{
    int i;
 
    // Map to store the frequency of
    // the numbers at the even indices
    map<int, int> te;
 
    // Map to store the frequency of
    // the numbers at the odd indices
    map<int, int> to;
 
    for (i = 0; i < n; i++)
    {
 
        // Checking if the index
        // is odd or even
        if (i % 2 == 0)
 
            // If the number is already present then,
            // just increase the occurrence by 1
            te[a[i]]++;
        else
 
            // If the number is already present then,
            // just increase the occurrence by 1
            to[a[i]]++;
    }
 
    // To store the character with
    // maximum frequency in even indices.
    int me = -1;
 
    // To store the character with
    // maximum frequency in odd indices.
    int mo = -1;
 
    // To store the frequency of the
    // maximum occurring number in even indices.
    int ce = -1;
 
    // To store the frequency of the
    // maximum occurring number in odd indices.
    int co = -1;
 
    // Iterating over Map of even indices to
    // get the maximum occurring number.
    for (auto it : te)
    {
        if (it.second > ce)
        {
            ce = it.second;
            me = it.first;
        }
    }
 
    // Iterating over Map of odd indices to
    // get the maximum occurring number.
    for (auto it : to)
    {
        if (it.second > co)
        {
            co = it.second;
            mo = it.first;
        }
    }
 
    // To store the final answer
    int res = 0;
 
    for (i = 0; i < n; i++)
    {
        if (i % 2 == 0)
        {
 
            // If the index is even but
            // a[i] != me
            // then a[i] needs to be replaced
            if (a[i] != me) res++;
        }
         
        else
        {
 
            // If the index is odd but
            // a[i] != mo
            // then a[i] needs to be replaced
            if (a[i] != mo) res++;
        }
    }
    return res;
}
 
// Driver Code
int main()
{
    int n = 4;
    int a[] = {3, 1, 3, 2};
    cout << minReplace(a, n) << endl;
    return 0;
}
 
// This code is contributed by
// sanjeev2552

Java




// Java implementation of the approach
import java.util.HashMap;
class GFG {
 
    // Function to return the minimum replacements
    public static int minReplace(int a[], int n)
    {
        int i;
 
        // Map to store the frequency of
        // the numbers at the even indices
        HashMap<Integer, Integer> te = new HashMap<>();
 
        // Map to store the frequency of
        // the numbers at the odd indices
        HashMap<Integer, Integer> to = new HashMap<>();
 
        for (i = 0; i < n; i++) {
 
            // Checking if the index
            // is odd or even
            if (i % 2 == 0) {
 
                // If the number is already present then,
                // just increase the occurrence by 1
                if (te.containsKey(a[i]))
                    te.put(a[i], te.get(a[i]) + 1);
                else
                    te.put(a[i], 1);
            }
            else {
 
                // If the number is already present then,
                // just increase the occurrence by 1
                if (to.containsKey(a[i]))
                    to.put(a[i], to.get(a[i]) + 1);
                else
                    to.put(a[i], 1);
            }
        }
 
        // To store the character with
        // maximum frequency in even indices.
        int me = -1;
 
        // To store the character with
        // maximum frequency in odd indices.
        int mo = -1;
 
        // To store the frequency of the
        // maximum occurring number in even indices.
        int ce = -1;
 
        // To store the frequency of the
        // maximum occurring number in odd indices.
        int co = -1;
 
        // Iterating over Map of even indices to
        // get the maximum occurring number.
        for (Integer It : te.keySet()) {
            if (te.get(It) > ce) {
                ce = te.get(It);
                me = It;
            }
        }
 
        // Iterating over Map of odd indices to
        // get the maximum occurring number.
        for (Integer It : to.keySet()) {
            if (to.get(It) > co) {
                co = to.get(It);
                mo = It;
            }
        }
 
        // To store the final answer
        int res = 0;
 
        for (i = 0; i < n; i++) {
            if (i % 2 == 0) {
 
                // If the index is even but
                // a[i] != me
                // then a[i] needs to be replaced
                if (a[i] != me)
                    res++;
            }
            else {
 
                // If the index is odd but
                // a[i] != mo
                // then a[i] needs to be replaced
                if (a[i] != mo)
                    res++;
            }
        }
 
        return res;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n;
        n = 4;
        int a[] = { 3, 1, 3, 2 };
        System.out.println(minReplace(a, n));
    }
}

Python3




# Python3 implementation of the approach
 
# Function to return the minimum replacements
def minReplace(a: list, n) -> int:
 
    # Map to store the frequency of
    # the numbers at the even indices
    te = dict()
 
    # Map to store the frequency of
    # the numbers at the odd indices
    to = dict()
 
    for i in range(n):
 
        # Checking if the index
        # is odd or even
        if i % 2 == 0:
 
            # If the number is already present then,
            # just increase the occurrence by 1
            if a[i] not in te:
                te[a[i]] = 1
            else:
                te[a[i]] += 1
        else:
 
            # If the number is already present then,
            # just increase the occurrence by 1
            if a[i] not in to:
                to[a[i]] = 1
            else:
                to[a[i]] += 1
 
    # To store the character with
    # maximum frequency in even indices.
    me = -1
 
    # To store the character with
    # maximum frequency in odd indices.
    mo = -1
 
    # To store the frequency of the
    # maximum occurring number in even indices.
    ce = -1
 
    # To store the frequency of the
    # maximum occurring number in odd indices.
    co = -1
 
    # Iterating over Map of even indices to
    # get the maximum occurring number.
    for it in te:
        if te[it] > ce:
            ce = te[it]
            me = it
 
    # Iterating over Map of odd indices to
    # get the maximum occurring number.
    for it in to:
        if to[it] > co:
            co = to[it]
            mo = it
 
    # To store the final answer
    res = 0
 
    for i in range(n):
        if i % 2 == 0:
 
            # If the index is even but
            # a[i] != me
            # then a[i] needs to be replaced
            if a[i] != me:
                res += 1
        else:
 
            # If the index is odd but
            # a[i] != mo
            # then a[i] needs to be replaced
            if a[i] != mo:
                res += 1
 
    return res
 
# Driver Code
if __name__ == "__main__":
    n = 4
    a = [3, 1, 3, 2]
    print(minReplace(a, n))
 
# This code is contributed by
# sanjeev2552

C#




// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // Function to return the minimum replacements
    public static int minReplace(int []a, int n)
    {
        int i;
 
        // Map to store the frequency of
        // the numbers at the even indices
        Dictionary<int,
                   int> te = new Dictionary<int,
                                            int>();
 
        // Map to store the frequency of
        // the numbers at the odd indices
        Dictionary<int,
                   int> to = new Dictionary<int,
                                            int>();
 
        for (i = 0; i < n; i++)
        {
 
            // Checking if the index
            // is odd or even
            if (i % 2 == 0)
            {
 
                // If the number is already present then,
                // just increase the occurrence by 1
                if (te.ContainsKey(a[i]))
                    te[a[i]] = te[a[i]] + 1;
                else
                    te.Add(a[i], 1);
            }
            else
            {
 
                // If the number is already present then,
                // just increase the occurrence by 1
                if (to.ContainsKey(a[i]))
                    to[a[i]] = te[a[i]] + 1;
                else
                    to.Add(a[i], 1);
            }
        }
 
        // To store the character with
        // maximum frequency in even indices.
        int me = -1;
 
        // To store the character with
        // maximum frequency in odd indices.
        int mo = -1;
 
        // To store the frequency of the
        // maximum occurring number in even indices.
        int ce = -1;
 
        // To store the frequency of the
        // maximum occurring number in odd indices.
        int co = -1;
 
        // Iterating over Map of even indices to
        // get the maximum occurring number.
        foreach (int It in te.Keys)
        {
            if (te[It] > ce)
            {
                ce = te[It];
                me = It;
            }
        }
 
        // Iterating over Map of odd indices to
        // get the maximum occurring number.
        foreach (int It in to.Keys)
        {
            if (to[It] > co)
            {
                co = to[It];
                mo = It;
            }
        }
 
        // To store the final answer
        int res = 0;
 
        for (i = 0; i < n; i++)
        {
            if (i % 2 == 0)
            {
 
                // If the index is even but
                // a[i] != me
                // then a[i] needs to be replaced
                if (a[i] != me)
                    res++;
            }
            else
            {
 
                // If the index is odd but
                // a[i] != mo
                // then a[i] needs to be replaced
                if (a[i] != mo)
                    res++;
            }
        }
        return res;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int n;
        n = 4;
        int []a = { 3, 1, 3, 2 };
        Console.WriteLine(minReplace(a, n));
    }
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to return the minimum replacements
function minReplace(a, n)
{
    var i;
 
    // Map to store the frequency of
    // the numbers at the even indices
    var te = new Map();
 
    // Map to store the frequency of
    // the numbers at the odd indices
    var to = new Map();
 
    for (i = 0; i < n; i++)
    {
 
        // Checking if the index
        // is odd or even
        if (i % 2 == 0)
 
            // If the number is already present then,
            // just increase the occurrence by 1
            if(te.has(a[i]))
                te.set(a[i], te.get(a[i])+1)
            else
                te.set(a[i],1)
        else
 
            // If the number is already present then,
            // just increase the occurrence by 1
            if(to.has(a[i]))
                to.set(a[i], to.get(a[i])+1)
            else
                to.set(a[i],1)
    }
 
    // To store the character with
    // maximum frequency in even indices.
    var me = -1;
 
    // To store the character with
    // maximum frequency in odd indices.
    var mo = -1;
 
    // To store the frequency of the
    // maximum occurring number in even indices.
    var ce = -1;
 
    // To store the frequency of the
    // maximum occurring number in odd indices.
    var co = -1;
 
    // Iterating over Map of even indices to
    // get the maximum occurring number.
    te.forEach((value, key) => {
         
        if (value > ce)
        {
            ce = value;
            me = key;
        }
    });
 
    // Iterating over Map of odd indices to
    // get the maximum occurring number.
    to.forEach((value, key) => {
        
        if (value > co)
        {
            co = value;
            mo = key;
        }
    });
 
    // To store the final answer
    var res = 0;
 
    for (i = 0; i < n; i++)
    {
        if (i % 2 == 0)
        {
 
            // If the index is even but
            // a[i] != me
            // then a[i] needs to be replaced
            if (a[i] != me)
              res++;
        }
         
        else
        {
 
            // If the index is odd but
            // a[i] != mo
            // then a[i] needs to be replaced
            if (a[i] != mo)
              res++;
        }
    }
    return res;
}
 
// Driver Code
var n = 4;
var a = [3, 1, 3, 2];
document.write( minReplace(a, n) );
 
</script>
Output: 
1

 




My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!