LeetCode 1187: Make Array Strictly Increasing Solution

Master LeetCode problem 1187 (Make Array Strictly Increasing), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1187. Make Array Strictly Increasing

Hard

Array Binary Search Dynamic Programming Sorting

Problem Explanation

Explanation:

To solve this problem, we can use dynamic programming. We will create a 2D dp array to keep track of the minimum number of operations needed to make a subarray of arr1 strictly increasing up to index i. We will iterate through both arr1 and arr2 and update our dp array accordingly.

At each index i of arr1, we have two choices:

If arr1[i] > arr1[i-1], we can keep arr1[i] as is without any operations.
If arr1[i] <= arr1[i-1], we need to find a suitable value from arr2 to replace arr1[i] to make it strictly increasing. We check all possible values in arr2 and choose the one that minimizes the number of operations.

After filling the dp array, we will find the minimum operations needed to make the entire arr1 strictly increasing by considering the last index of the dp array.

Time Complexity:

The time complexity of this approach is O(n^2), where n is the length of arr1.

Space Complexity:

The space complexity is O(n), where n is the length of arr1.

Solution Code

class Solution {
    public int makeArrayIncreasing(int[] arr1, int[] arr2) {
        Arrays.sort(arr2);
        int n = arr1.length;
        int[][] dp = new int[n + 1][n + 1];
        for (int i = 0; i <= n; i++) {
            Arrays.fill(dp[i], Integer.MAX_VALUE);
        }
        dp[0][0] = -1;
        
        for (int i = 1; i <= n; i++) {
            int prevVal = dp[i - 1][0];
            for (int j = 0; j <= n; j++) {
                if (arr1[i - 1] > prevVal) {
                    dp[i][j] = Math.min(dp[i][j], dp[i - 1][j]);
                }
                if (j > 0) {
                    int replaceIdx = Arrays.binarySearch(arr2, prevVal + 1);
                    if (replaceIdx < 0) {
                        replaceIdx = -(replaceIdx + 1);
                    }
                    if (replaceIdx < arr2.length) {
                        dp[i][j] = Math.min(dp[i][j], dp[i - 1][j - 1] + 1);
                    }
                }
                if (j < n) {
                    prevVal = arr1[i - 1];
                }
            }
        }
        
        int res = Integer.MAX_VALUE;
        for (int j = 0; j <= n; j++) {
            if (dp[n][j] != Integer.MAX_VALUE) {
                res = Math.min(res, dp[n][j]);
            }
        }

        return res == Integer.MAX_VALUE ? -1 : res;
    }
}

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