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Minimum Operations to Make the Array K-Increasing

Difficulty: Hard

Problem Description

You are given a 0-indexed array arr consisting of n positive integers, and a positive integer k. The array arr is called K-increasing if arr[i-k] <= arr[i] holds for every index i, where k <= i <= n-1. In one operation, you can choose an index i and change arr[i] into any positive integer. Return the minimum number of operations required to make the array K-increasing for the given k.

Key Insights

  • The problem can be divided into k separate subsequences based on the index modulo k.
  • For each subsequence, we need to find the minimum number of operations required to make it non-decreasing.
  • The problem of making a subsequence non-decreasing can be solved using the concept of Longest Increasing Subsequence (LIS).
  • The number of operations needed for each subsequence is the difference between its length and the length of its LIS.

Space and Time Complexity

Time Complexity: O(n log n)
Space Complexity: O(n)

Solution

To solve the problem, we can break the array into k subsequences based on their indices. For each subsequence, we will identify the longest increasing subsequence (LIS) using binary search, which allows us to efficiently compute the number of operations needed to make the subsequence non-decreasing. The total number of operations required will be the sum of operations for all k subsequences.

Code Solutions

def min_operations(arr, k):
    from bisect import bisect_left
    
    def lis_length(subseq):
        dp = []

        for num in subseq:
            pos = bisect_left(dp, num)
            if pos == len(dp):
                dp.append(num)
            else:
                dp[pos] = num

        return len(dp)

    n = len(arr)
    operations = 0

    for i in range(k):
        subseq = [arr[j] for j in range(i, n, k)]
        operations += len(subseq) - lis_length(subseq)

    return operations
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