# Solving Leetcode Interviews in Seconds with AI: Make Array Strictly Increasing


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1187" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Given two integer arrays arr1 and arr2, return the minimum number of operations (possibly zero) needed to make arr1 strictly increasing. In one operation, you can choose two indices 0 <= i < arr1.length and 0 <= j < arr2.length and do the assignment arr1[i] = arr2[j]. If there is no way to make arr1 strictly increasing, return -1.   Example 1:  Input: arr1 = [1,5,3,6,7], arr2 = [1,3,2,4] Output: 1 Explanation: Replace 5 with 2, then arr1 = [1, 2, 3, 6, 7].  Example 2:  Input: arr1 = [1,5,3,6,7], arr2 = [4,3,1] Output: 2 Explanation: Replace 5 with 3 and then replace 3 with 4. arr1 = [1, 3, 4, 6, 7].  Example 3:  Input: arr1 = [1,5,3,6,7], arr2 = [1,6,3,3] Output: -1 Explanation: You can't make arr1 strictly increasing.   Constraints:  1 <= arr1.length, arr2.length <= 2000 0 <= arr1[i], arr2[i] <= 10^9    

	# Explanation
	Here's the breakdown of the approach and the Python code:

*   **High-Level Approach:**
    *   Dynamic Programming: Use a DP table `dp[i][j]` where `dp[i][j]` represents the minimum number of operations to make `arr1[0...i]` strictly increasing, and `arr1[i]` is equal to the `j`-th smallest element from `arr2` (or `float('inf')` if no replacement happens).
    *   Binary Search Optimization: To efficiently find a suitable replacement from `arr2` that is greater than the previous element in `arr1`, use binary search on the sorted `arr2`.
    *   Base Case and Transitions: Initialize the base case where the first element of `arr1` might need replacement. Then, for each subsequent element, determine whether a replacement is needed and calculate the minimum operations accordingly.

*   **Complexity:**
    *   Runtime: O(n * m * log(m)), where n is the length of `arr1` and m is the length of `arr2`.  Sorting `arr2` takes O(m log m), and the DP part takes O(n * m * log(m)).
    *   Storage: O(n * m) due to the `dp` table.

	
	# Code
	```python
	def makeArrayIncreasing(arr1, arr2):
    arr2 = sorted(list(set(arr2)))
    n = len(arr1)
    m = len(arr2)

    dp = {}  # dp[(index, prev_val)] = min_ops

    def solve(index, prev_val):
        if index == n:
            return 0

        if (index, prev_val) in dp:
            return dp[(index, prev_val)]

        ans = float('inf')

        # Option 1: No replacement
        if arr1[index] > prev_val:
            ans = min(ans, solve(index + 1, arr1[index]))

        # Option 2: Replace arr1[index] with an element from arr2
        l, r = 0, m - 1
        replace_idx = -1
        while l <= r:
            mid = (l + r) // 2
            if arr2[mid] > prev_val:
                replace_idx = mid
                r = mid - 1
            else:
                l = mid + 1

        if replace_idx != -1:
            ans = min(ans, 1 + solve(index + 1, arr2[replace_idx]))

        dp[(index, prev_val)] = ans
        return ans

    result = solve(0, -1)
    return result if result != float('inf') else -1
	```