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.

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 :

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// 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));
    }
}

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Output:

1


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