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Minimal price to make even Array

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Given an enter array, the duty is to transform the given array into an ‘even array’. A good array is an array the place each quantity is a fair quantity. For making an array even you are able to do these two operations:

  • If the present quantity is odd, then calculate the worth to make the quantity even. For making a quantity even, you may swap adjoining positions with one place as even and one other as odd.
    Price = the whole variety of swaps made.
  • If you wish to delete the present aspect from the array then the associated fee shall be 5.


Enter: {1243, 267, 2315}
Output: 5
Clarification: For making 1243 a fair quantity, we’ll swap 4 and three. This can price 1 operation.
For making 267 a fair quantity, we’ll swap 6 and seven. This can price 1 operation.
For making 2315 a fair quantity, we’ll swap 2 and three adopted by 2 and 1, after which 2 and 5. This can price 3 operations.
So the minimal price shall be 5.

Enter: {12, 23, 2171315}
Output: 7
Clarification: 12 is already a fair quantity so it should price 0 operations. 
For making 23 a fair quantity, we’ll swap 2 and three. This can price 1 operation.
For making 2171315 a fair quantity, we have to carry out 6 operations so as an alternative of creating it even we’ll delete it from the array which can price 5 operations.
So after deleting, the minimal price shall be 6.

Strategy: To resolve the issue observe the under concept:

The thought is to take care of a variable for calculating the minimal price and for each single aspect of the array select the minimal between 5 or the whole variety of swaps required.

This whole factor might be executed by following the below-mentioned steps:

  • Implement a operate named makeEven which takes an integer as an enter and returns the variety of swaps required for making that quantity even or -1 if there is no such thing as a even digit in that quantity.
  • Initialize a variable minCost for storing the minimal price and iterate over the enter array.
  • For each single iteration, Initialize a variable price = makeEven ( currentElement). 
  • if price == -1 then add 5 to minCost in any other case add minimal amongst 5 and value to minCost.
  • return minCost.

Beneath is the implementation for the above strategy:


#embody <bits/stdc++.h>

utilizing namespace std;


int makeEven(int n)


    int price = 0;




    whereas (n % 2 != 0) {




        n /= 10;







    if (n == 0)

        return -1;

    return price;



int makeArrayEven(vector<int> arr)


    int minCost = 0;

    for (int i = 0; i < arr.measurement(); i++) {

        int price = makeEven(arr[i]);




        if (price != -1)

            minCost += min(5, price);





            minCost += 5;







    return minCost;



int major()


    vector<int> arr{ 12, 23, 2171315 };



    cout << makeArrayEven(arr);


    return 0;


Time Complexity: O(N), the place  N is the dimensions of the array,
Auxiliary Area: O(1)



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