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Minimum Operations to Make the Array K-Increasing

Number: 2234

Difficulty: Hard

Paid? No

Companies: Amazon

Problem Description

Given a 0-indexed array arr of n positive integers and a positive integer k, an array is called K-increasing if for every index i (where k ≤ i < n) the condition arr[i-k] <= arr[i] holds. In one operation, you can change any arr[i] to any positive integer. Return the minimum number of operations required to make the array K-increasing.

Key Insights

  • The K-increasing condition only applies between elements whose indices are k apart.
  • Partition the array into k independent subsequences where the jth subsequence consists of elements arr[j], arr[j+k], arr[j+2k], ….
  • Each subsequence must be made non-decreasing.
  • The problem of minimizing operations in each subsequence equates to finding the Longest Non-Decreasing Subsequence (LNDS) within that subsequence; elements not part of the LNDS will need to be changed.
  • The total number of operations is the sum over all subsequences of (length of subsequence - LNDS length).

Space and Time Complexity

Time Complexity: O(n log n) – For each of the k subsequences, we perform a binary search-based LNDS computation. Space Complexity: O(n) – Extra space is used to store the subsequences and temporary arrays during LNDS computation.

Solution

We start by dividing the array into k independent subsequences. Each subsequence is processed to determine how many elements must be replaced to form a non-decreasing sequence. This is accomplished by computing the Longest Non-Decreasing Subsequence (LNDS) for that subsequence using a greedy algorithm with binary search (similar to patience sorting). For each subsequence, the minimum changes required is the difference between its length and the LNDS length. Summing these differences across all k subsequences yields the final answer.

Data Structures and Algorithms Used:

  • Arrays/lists to hold subsequences.
  • Greedy algorithm with binary search (using a temporary list to simulate the LNDS formation) for computing LNDS in O(n log n) time.

This approach takes advantage of the fact that modifications in one subsequence do not affect the others, allowing independent processing.

Code Solutions

# Python Code Solution
import bisect

def kIncreasing(arr, k):
    # Helper function to calculate the length of longest non-decreasing subsequence in a given sequence.
    def longest_nondecreasing_subsequence(seq):
        # This list will store the current non-decreasing sequence (ends values)
        temp = []
        for num in seq:
            # Use bisect_right for non-decreasing (<= allowed) instead of strictly increasing
            pos = bisect.bisect_right(temp, num)
            if pos < len(temp):
                temp[pos] = num  # replace to maintain lower tail values
            else:
                temp.append(num)
        return len(temp)
    
    operations = 0
    # Process each subsequence starting at index 0, 1, ..., k-1
    for i in range(k):
        subseq = []
        # Build the subsequence for the current remainder i when divided by k
        for j in range(i, len(arr), k):
            subseq.append(arr[j])
        # The number of operations is the change required to make subsequence non-decreasing.
        lnds = longest_nondecreasing_subsequence(subseq)
        operations += (len(subseq) - lnds)
    return operations

# Example usage:
print(kIncreasing([5,4,3,2,1], 1))  # Expected output: 4
print(kIncreasing([4,1,5,2,6,2], 2))  # Expected output: 0
print(kIncreasing([4,1,5,2,6,2], 3))  # Expected output: 2
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