Given an array A containing N non-negative integers. Only following operation can be performed on the array:
A[i] = ( A[i] + 1 ) % 3
where A[i] is the element of array A at index i, and performing this operation once costs 1 unit. Find the minimum cost to make all elements equal.
Examples :
Input : 1 0 3 2
Output : 4 units
Explanation: First 3 is converted to 1(3 + 1 % 3 = 1).
Then, if we try to make all elements equal to 1, then
converting 0 to 1 will cost "1" and 2 to 1 will cost "2".
Therefore, in total cost=3 + 1(to convert 3 to 1) = 4,
which is minimum.
Input : 98 4 3 1
Output : 6 units
Explanation: 98, 4 and 3 are converted to 0, 2 and 1.
So, now array becomes {0, 2, 1, 1}. If we try to convert
every element in our new array equal to 1, then converting
0 to 1 will cost "1" and 2 to 1 will cost "2". So, total
cost= 3+ 3(conversion of 98, 4 and 3) = 6, which is the minimum cost.
Approach: At first, we will try to convert all the numbers greater than 2 to a number between 0 to 2.Since, numbers from 0 to 2 will not exceed 2 using the only allowed operation A(i)=(A(i)+1) % 3(No matter how many times we do this operation). Then we will fix one of the 3 possible values(0 to 2) and find the cost of making all elements equal to it.
Minimum cost out of the three + the extra cost to convert all numbers greater than 2 to a number between 0 to 2 will be our answer.One special case we need look here is, if all the elements are equal by default (like in array {7, 7, 7, 7}), then our answer will be zero.
C++
// CPP program to find minimum cost to // equalize an array with increment under // modulo 3 operation. #include <bits/stdc++.h> using namespace std; int mincost(int a[], int n) { int c = 1; // loop to check whether all // elements are equal or not for (int i = 0; i < n; i++) if (a[i] == a[i + 1]) c++; // A special case when all // elements are equal if (c == n) return 0; // variable that counts total // numbers greater 2 int x = 0; // loop to convert all numbers greater // than 2 to a range between 0 to 2. for (int i = 0; i < n; i++) { if (a[i] > 2) { // number greater than 2 gets // converted to a number between 0 to 2. a[i] = (a[i] + 1) % 3; x += 1; } } // variables to count 3 possible ways // to make all elements equal after // reducing them to range [0, 2] int c0 = 0, c1 = 0, c2 = 0; // loop that counts total cost to // make all elements equal to 0. for (int i = 0; i < n; i++) { if (a[i] == 1) { // since we will have to use the // operation 2 times to convert 1 to 0 // (1+1)%3=2 and then (2+1)%3=0 c0 += 2; } // here we use it once since (2+1%3=0) else if (a[i] == 2) c0 += 1; } // loop that counts total cost to // make all elements equal to 1 for (int i = 0; i < n; i++) { // since (2+1%3=0) and (0+1%3=1) if (a[i] == 2) c1 += 2; // since (0+1%3=1) else if (a[i] == 0) c1 += 1; } // loop that counts total cost // to make all elements equal to 2 for (int i = 0; i < n; i++) { // since 0+1%3=1 and 1+1%3=2 if (a[i] == 0) c2 += 2; else if (a[i] == 1) c2 += 1; } // finally the one with minimum // cost will be our answer return x + min({ c0, c1, c2 }); } // Driver program to run above function int main() { int a[] = { 98, 4, 3, 1 }; int n = sizeof(a)/sizeof(a[0]); cout << mincost(a, n)<<" units"; return 0; } |
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Java
// Java program to find minimum // cost to equalize an array // with increment under // modulo 3 operation. import java.io.*; class GFG { // Function to find minimum cost static int mincost(int a[], int n) { int c = 1; // loop to check whether all // elements are equal or not for (int i = 0; i < n - 1; i++) if (a[i] == a[i + 1]) c++; // A special case when all // elements are equal if (c == n) return 0; // variable that counts total // numbers greater 2 int x = 0; // loop to convert all numbers // greater than 2 to a range // between 0 to 2. for (int i = 0; i < n - 1; i++) { if (a[i] > 2) { // number greater than 2 // gets converted to a // number between 0 to 2. a[i] = (a[i] + 1) % 3; x += 1; } } // variables to count 3 possible // ways to make all elements equal // after reducing them to range [0, 2] int c0 = 0, c1 = 0, c2 = 0; // loop that counts total cost to // make all elements equal to 0. for (int i = 0; i < n - 1; i++) { if (a[i] == 1) { // since we will have to // use the operation 2 // times to convert 1 to 0 // (1+1)%3=2 and then (2+1)%3=0 c0 += 2; } // here we use it // once since (2+1%3=0) else if (a[i] == 2) c0 += 1; } // loop that counts total cost to // make all elements equal to 1 for (int i = 0; i < n - 1; i++) { // since (2+1%3=0) // and (0+1%3=1) if (a[i] == 2) c1 += 2; // since (0+1%3=1) else if (a[i] == 0) c1 += 1; } // loop that counts total cost // to make all elements equal to 2 for (int i = 0; i < n - 1; i++) { // since 0+1%3=1 // and 1+1%3=2 if (a[i] == 0) c2 += 2; else if (a[i] == 1) c2 += 1; } // finally the one with minimum // cost will be our answer return x + Math.min(c0, Math.min(c1, c2)); } // Driver Code public static void main (String[] args) { int a[] = new int[]{98, 4, 3, 1}; int n = a.length; System.out.println(mincost(a, n) + " " + "units"); } } // This code is contributed by ajit |
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Python3
# Python code to illustrate above approach # function to calculate minimum cost def mincost(a, n): # loop to check whether all elements # are equal or not c = 1 for i in range(0, n-1): if (a[i] == a[i + 1]): c += 1 # A special case when all elements # are equal if (c == n): return 0 # loop to count total no of conversion # of numbers greater than 2 to a # number between 0 to 2. x = 0 for i in range(n): if a[i]>2: a[i]=(a[i]+1)% 3 x += 1 # variables to count 3 possible ways # to make all elements equal after # reducing them to range [0, 2] c0 = c1 = c2 = 0 # loop that counts total cost to # make all elements equal to 0. for i in a: if (i == 1): c0+= 2 elif (i == 2): c0+= 1 # loop that counts total cost to # make all elements equal to 1. for i in a: if (i == 0): c1+= 1 elif (i == 2): c1+= 2 # loop that counts total cost to # make all elements equal to 2. for i in a: if (i == 0): c2+= 2 elif (i == 1): c2+= 1 # finally the one with minimum cost # plus the extra cost to convert numbers # greater than 2 will be our answer return min(c1, c2, c0)+x # Driver code n = 4a = [98, 4, 3, 1] c = 1print(mincost(a, n), "units") |
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C#
// C# program to find minimum cost to // equalize an array with increment // under modulo 3 operation. using System; class GFG { // Function to find minimum cost static int mincost(int []a, int n) { int c = 1; // loop to check whether all // elements are equal or not for (int i = 0; i < n - 1; i++) if (a[i] == a[i + 1]) c++; // A special case when all // elements are equal if (c == n) return 0; // variable that counts total // numbers greater 2 int x = 0; // loop to convert all numbers greater // than 2 to a range between 0 to 2. for (int i = 0; i < n - 1; i++) { if (a[i] > 2) { // number greater than 2 gets // converted to a number // between 0 to 2. a[i] = (a[i] + 1) % 3; x += 1; } } // variables to count 3 possible ways // to make all elements equal after // reducing them to range [0, 2] int c0 = 0, c1 = 0, c2 = 0; // loop that counts total cost to // make all elements equal to 0. for (int i = 0; i < n - 1; i++) { if (a[i] == 1) { // since we will have to use the // operation 2 times to convert 1 to 0 // (1+1)%3=2 and then (2+1)%3=0 c0 += 2; } // here we use it once since (2+1%3=0) else if (a[i] == 2) c0 += 1; } // loop that counts total cost to // make all elements equal to 1 for (int i = 0; i < n - 1; i++) { // since (2+1%3=0) and (0+1%3=1) if (a[i] == 2) c1 += 2; // since (0+1%3=1) else if (a[i] == 0) c1 += 1; } // loop that counts total cost // to make all elements equal to 2 for (int i = 0; i < n - 1; i++) { // since 0+1%3=1 and 1+1%3=2 if (a[i] == 0) c2 += 2; else if (a[i] == 1) c2 += 1; } // finally the one with minimum // cost will be our answer return x + Math.Min(c0, Math.Min(c1, c2) ); } // Driver Code public static void Main() { int []a = new int[]{98, 4, 3, 1}; int n = a.Length; Console.Write(mincost(a, n) + " " + "units"); } } // This code is contributed by Sam007. |
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PHP
<?php // PHP program to find minimum // cost to equalize an array // with increment under modulo // 3 operation. function mincost($a, $n) { $c = 1; // loop to check whether all // elements are equal or not for ($i = 0; $i < $n; $i++) $c++; // A special case when all // elements are equal if ($c == $n) return 0; // variable that counts // total numbers greater 2 $x = 0; // loop to convert all // numbers greater than 2 // to a range between 0 to 2. for ($i = 0; $i < $n; $i++) { if ($a[$i] > 2) { // number greater than 2 // gets converted to a // number between 0 to 2. $a[$i] = ($a[$i] + 1) % 3; $x += 1; } } // variables to count 3 // possible ways to make // all elements equal after // reducing them to range [0, 2] $c0 = 0; $c1 = 0; $c2 = 0; // loop that counts total // cost to make all elements // equal to 0. for ($i = 0; $i < $n; $i++) { if ($a[$i] == 1) { // since we will have to // use the operation 2 // times to convert 1 to 0 // (1+1)%3=2 and then (2+1)%3=0 $c0 += 2; } // here we use it // once since (2+1%3=0) else if ($a[$i] == 2) $c0 += 1; } // loop that counts total // cost to make all // elements equal to 1 for ($i = 0; $i < $n; $i++) { // since (2+1%3=0) and // (0+1%3=1) if ($a[$i] == 2) $c1 += 2; // since (0+1%3=1) else if ($a[$i] == 0) $c1 += 1; } // loop that counts total // cost to make all // elements equal to 2 for ($i = 0; $i < $n; $i++) { // since 0+1%3=1 // and 1+1%3=2 if ($a[$i] == 0) $c2 += 2; else if ($a[$i] == 1) $c2 += 1; } // finally the one with // minimum cost will // be our answer return $x + min($c0, $c1, $c2 ); } // Driver Code $a = array(98, 4, 3, 1); $n = sizeof($a); echo mincost($a, $n), " units"; // This code is contributed by ajit ?> |
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Output :
6 units
Time Complexity :O(n)
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