How do you solve Make Array Strictly Increasing on LeetCode?

Make Array Strictly Increasing (LeetCode #1187) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m) time complexity and O(n) space complexity. Key insight: Dynamic programming can reduce the complexity of problems involving sequences.

What is the brute force solution for Make Array Strictly Increasing?

The brute force approach for Make Array Strictly Increasing is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves trying every possible combination of replacements from arr2 into arr1 to see if we can make arr1 strictly increasing. This method is straightforward but inefficient, as it checks all possibilities. The optimal Optimal Solution approach improves this to O(n * m).

What algorithm or data structure is used to solve Make Array Strictly Increasing?

Make Array Strictly Increasing uses the following concepts: Array, Binary Search, Dynamic Programming, Sorting. The recommended patterns to study are: Dynamic Programming, Sorting, Array.

What are the interview tips for Make Array Strictly Increasing?

Explain your thought process clearly when discussing your approach. Consider edge cases and constraints upfront to avoid surprises later. Practice dynamic programming problems to become comfortable with state transitions.

What are common mistakes when solving Make Array Strictly Increasing?

Not considering the sorted order of arr2 when replacing elements. Failing to check edge cases where arr1 is already strictly increasing.

#1187

Make Array Strictly Increasing

Hard

ArrayBinary SearchDynamic ProgrammingSortingDynamic ProgrammingSortingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m)
Space	O(1)	O(n)

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Intuition

Time O(n * m)Space O(n)

The optimal approach uses dynamic programming to efficiently track the minimum operations needed to make arr1 strictly increasing by leveraging sorted arr2. We maintain a state that considers the last used element from arr2 to ensure we can build a strictly increasing sequence.

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Algorithm

4 steps

1Step 1: Sort arr2 and remove duplicates to simplify replacements.
2Step 2: Initialize a DP array where dp[i] represents the minimum operations needed to make arr1[0..i] strictly increasing.
3Step 3: For each element in arr1, check against all elements in arr2 to find valid replacements that maintain the strictly increasing order.
4Step 4: Update the DP array based on the minimum operations found.

solution.py16 lines

1# Full working Python code
2import bisect
3
4def min_operations(arr1, arr2):
5    arr2 = sorted(set(arr2))
6    n = len(arr1)
7    dp = [float('inf')] * (n + 1)
8    dp[0] = 0  # No operations needed for an empty array
9
10    for i in range(1, n + 1):
11        for j in range(len(arr2)):
12            if arr2[j] > (arr1[i - 1] if i > 1 else float('-inf')):
13                dp[i] = min(dp[i], dp[i - 1])
14            if j > 0 and arr2[j] > arr1[i - 1]:
15                dp[i] = min(dp[i], dp[i - 1] + 1)
16    return dp[n] if dp[n] != float('inf') else -1

ℹ

Complexity note: The time complexity is O(n * m) where n is the length of arr1 and m is the length of arr2. This is due to the nested loops iterating through arr1 and arr2.

1Dynamic programming can reduce the complexity of problems involving sequences.
2Sorting and removing duplicates can simplify the problem space.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.