1187

1187. Make Array Strictly Increasing (Hard)

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

 

Related Topics:
Dynamic Programming

Solution 1. DP Top-down

// OJ: https://leetcode.com/problems/make-array-strictly-increasing/
// Author: github.com/lzl124631x
// Time: O(MN * logN)
// Space: O(MN)
// Ref: https://leetcode.com/problems/make-array-strictly-increasing/discuss/379095/C%2B%2B-DFS-%2B-Memo
class Solution {
    int dp[2001][2001] = {};
    int dfs(vector<int>& A, vector<int>& B, int i, int prev) {
        if (i >= A.size()) return 1;
        int j = upper_bound(B.begin(), B.end(), prev) - B.begin();
        if (dp[i][j]) return dp[i][j];
        int skip = prev < A[i] ? dfs(A, B, i + 1, A[i]) : B.size() + 1;
        int swap = j < B.size() ? 1 + dfs(A, B, i + 1, B[j]) : B.size() + 1;
        return dp[i][j] = min(skip, swap);
    }
public:
    int makeArrayIncreasing(vector<int>& A, vector<int>& B) {
        sort(B.begin(), B.end());
        auto ans = dfs(A, B, 0, INT_MIN);
        return ans > B.size() ? -1 : ans - 1;
    }
};

Or pass the previous j value into dfs so that j only traverse B in one pass.

// OJ: https://leetcode.com/problems/make-array-strictly-increasing/
// Author: github.com/lzl124631x
// Time: O(MN)
// Space: O(MN)
// Ref: https://leetcode.com/problems/make-array-strictly-increasing/discuss/379095/C%2B%2B-DFS-%2B-Memo
class Solution {
    int dp[2001][2001] = {};
    int dfs(vector<int>& A, vector<int>& B, int i, int j, int prev) {
        if (i >= A.size()) return 1;
        j = upper_bound(B.begin() + j, B.end(), prev) - B.begin();
        if (dp[i][j]) return dp[i][j];
        int skip = prev < A[i] ? dfs(A, B, i + 1, j, A[i]) : B.size() + 1;
        int swap = j < B.size() ? 1 + dfs(A, B, i + 1, j, B[j]) : B.size() + 1;
        return dp[i][j] = min(skip, swap);
    }
public:
    int makeArrayIncreasing(vector<int>& A, vector<int>& B) {
        sort(B.begin(), B.end());
        auto ans = dfs(A, B, 0, 0, INT_MIN);
        return ans > B.size() ? -1 : ans - 1;
    }
};

Solution 2. DP

// OJ: https://leetcode.com/problems/make-array-strictly-increasing/
// Author: github.com/lzl124631x
// Time: O(MN * logN)
// Space: O(MN)
// Ref: https://leetcode.com/problems/make-array-strictly-increasing/discuss/377403/Python-DP-solution-with-explanation.
class Solution {
public:
    int makeArrayIncreasing(vector<int>& A, vector<int>& B) {
        int M = A.size(), N = B.size();
        sort(B.begin(), B.end());
        unordered_map<int, int> dp;
        dp[-1] = 0;
        for (int a : A) {
            unordered_map<int, int> tmp;
            for (auto &p : dp) {
                if (a > p.first) {
                    int cnt = tmp.count(a) ? tmp[a] : INT_MAX;
                    tmp[a] = min(cnt, p.second);
                }
                int b = upper_bound(B.begin(), B.end(), p.first) - B.begin();
                if (b < N) {
                    int cnt = tmp.count(B[b]) ? tmp[B[b]] : INT_MAX;
                    tmp[B[b]] = min(cnt, p.second + 1);
                }
            }
            dp = tmp;
        }
        if (dp.empty()) return -1;
        int ans = INT_MAX;
        for (auto &p : dp) ans = min(ans, p.second);
        return ans;
    }
};