Make Array Strictly Increasing - Leetcode Solution

Problem Statement: Given two arrays arr1 and arr2, you have to convert arr1 into a strictly increasing array such that all its elements are distinct and arr1[i] > arr1[i-1]. In a single operation, you can choose any index i where 0 <= i < arr1.length and change arr1[i] to any integer. Additionally, you have to ensure that the resulting array arr1 is such that arr1[i] > arr2[i] for all 0 <= i < arr1.length.

Solution: The problem statement states that we need to change the original array arr1 such that it is strictly increasing, and arr1[i] > arr2[i] for all i.

We can solve this problem by using dynamic programming. Let's define a 2D array dp, where dp[i][j] is the minimum number of operations needed to make the first i elements of arr1 strictly increasing, and the ith element is arr1[i-1], and the jth element of arr2 is arr2[j-1].

We can use the following recurrence relation to fill the dp array:

if arr1[i-1] > arr1[i-2] and arr1[i-1] > arr2[j-1]: dp[i][j] = dp[i-1][j] if arr1[i-1] > arr1[i-2] and arr1[i-1] <= arr2[j-1]: dp[i][j] = min(dp[i][j], dp[i-1][j] + 1) if arr1[i-1] <= arr1[i-2] and arr1[i-1] > arr2[j-1]: dp[i][j] = min(dp[i][j], dp[i-1][j-1]) if arr1[i-1] <= arr1[i-2] and arr1[i-1] <= arr2[j-1]: dp[i][j] = inf

Where inf denotes a very large number, and dp[0][j] = 0 for all j.

The first case denotes that we do not have to change the ith element of arr1 if it is already strictly increasing and greater than arr2[j-1]. Thus, we can copy the minimum operations needed from the previous index, dp[i-1][j].

The second case denotes that we can change the ith element of arr1 to any number greater than arr1[i-2], and less than or equal to arr2[j-1], and add 1 to the minimum operations needed to make the first i elements strictly increasing.

The third case denotes that we can change the ith element of arr1 to any number greater than arr2[j-1], and the jth element of arr2 is not used in the current index, thus we can copy the minimum operations needed from dp[i-1][j-1].

The fourth case denotes that there is no way to make the ith element of arr1 strictly increasing and greater than arr2[j-1], thus we can set dp[i][j] to inf.

Finally, the minimum operations needed to make the entire array arr1 strictly increasing and greater than arr2 can be found by taking the minimum of dp[n][j], for all j.

Thus, the time complexity of our solution is O(nm), where n is the length of arr1, and m is the length of arr2, and the space complexity is also O(nm) for the dp array.

Code:

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