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

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 :

```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.
```

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

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 of that sequence will have to be replaced with the most occurring number in the same sequence.
• Finally, the count of the replacements from the previous step will be the answer.

Below is the implementation of the above approach :

## C++

 `// C++ implementation of the approach ` `#include ` `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 te = ``new` `HashMap<>(); ` ` `  `        ``// Map to store the frequency of ` `        ``// the numbers at the odd indices ` `        ``HashMap 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 `

Output:

```1
```

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