Given an array arr[] consisting of N integers and an integer K, the task is to find the minimum cost required to make each element of every subarray of length K equal. Cost of replacing any array element by another element is the absolute difference between the two.
Examples:
Input: A[] = {1, 2, 3, 4, 6}, K = 3
Output: 7
Explanation:
Subarray 1: Cost to convert subarray {1, 2, 3} to {2, 2, 2} = |1-2| + |2-2| + |3-2| = 2
Subarray 2: Cost to convert subarray {2, 3, 4} to {3, 3, 3} = |2-3| + |3-3| + |4-3| = 2
Subarray 3: Cost to convert subarray {3, 4, 6} to {4, 4, 4} = |3-4| + |4-4| + |6-4| = 3
Minimum Cost = 2 + 2 + 3 = 7/
Input: A[] = {2, 3, 4, 4, 1, 7, 6}, K = 4
Output: 21
Approach:
To find the minimum cost to convert each element of the subarray to a single element, change every element of the subarray to the median of that subarray. Follow the steps below to solve the problem:
To find the median for each running subarray efficiently, use a multiset to get the sorted order of elements in each subarray. Median will be the middle element of this multiset.
For the next subarray remove the leftmost element of the previous subarray from the multiset, add the current element to the multiset.
Keep a pointer mid to efficiently keep track of the middle element of the multiset.
If the newly added element is smaller than the previous middle element, move mid to its immediate smaller element. Otherwise, move mid to its immediate next element.
Calaculate cost of replacing every element of the subarray by the equation | A[i] – medianeach_subarray |.
Finally print the total cost.
Below is the implementation for the above approach:
C++
// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the minimum
// cost to convert each element of
// every subarray of size K equal
int minimumCost(vector arr, int n,
int k)
{
// Stores the minimum cost
int totalcost = 0;
int i, j;
// Stores the first K elements
multiset<int> mp(arr.begin(),
arr.begin() + k);
if (k == n) {
// Obtain the middle element of
// the multiset
auto mid = next(mp.begin(),
n / 2 - ((k + 1) % 2));
int z = *mid;
// Calculate cost for the subarray
for (i = 0; i < n; i++)
totalcost += abs(z - arr[i]);
// Return the total cost
return totalcost;
}
else {
// Obtain the middle element
// in multiset
auto mid = next(mp.begin(),
k / 2 - ((k + 1) % 2));
for (i = k; i < n; i++) {
int zz = *mid;
int cost = 0;
for (j = i - k; j < i; j++) {
// Cost for the previous
// k length subarray
cost += abs(arr[j] - zz);
}
totalcost += cost;
// Insert current element
// into multiset
mp.insert(arr[i]);
if (arr[i] < *mid) {
// New element appears
// to the left of mid
mid--;
}
if (arr[i - k] <= *mid) {
// New element appears
// to the right of mid
mid++;
}
// Remove leftmost element
// from the window
mp.erase(mp.lower_bound(arr[i - k]));
// For last element
if (i == n - 1) {
zz = *mid;
cost = 0;
for (j = i - k + 1;
j <= i; j++) {
// Calculate cost for the subarray
cost += abs(zz - arr[j]);
}
totalcost += cost;
}
}
// Return the total cost
return totalcost;
}
}
// Driver Code
int main()
{
int N = 5, K = 3;
vector<int> A({ 1, 2, 3, 4, 6 });
cout << minimumCost(A, N, K);
}
Output:
7
Time Complexity: O(NlogN)
Auxiliary Space: O(1)
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#arrays #greedy #mathematical #sorting #cpp-multiset #median-finding #sliding-window #subarray
