Form Largest Integer With Digits That Add up to Target

Description

Given an array of integers cost and an integer target, return the maximum integer you can paint under the following rules:

• The cost of painting a digit (i + 1) is given by cost[i] (0-indexed).
• The total cost used must be equal to target.
• The integer does not have 0 digits.

Since the answer may be very large, return it as a string. If there is no way to paint any integer given the condition, return "0".

Example 1:

Input: cost = [4,3,2,5,6,7,2,5,5], target = 9 Output: “7772” Explanation: The cost to paint the digit ‘7’ is 2, and the digit ‘2’ is 3. Then cost(“7772”) = 2*3+ 3*1 = 9. You could also paint “977”, but “7772” is the largest number. Digit cost 1 → 4 2 → 3 3 → 2 4 → 5 5 → 6 6 → 7 7 → 2 8 → 5 9 → 5

Example 2:

Input: cost = [7,6,5,5,5,6,8,7,8], target = 12 Output: “85” Explanation: The cost to paint the digit ‘8’ is 7, and the digit ‘5’ is 5. Then cost(“85”) = 7 + 5 = 12.

Example 3:

Input: cost = [2,4,6,2,4,6,4,4,4], target = 5 Output: “0” Explanation: It is impossible to paint any integer with total cost equal to target.

Constraints:

• cost.length == 9
• 1 <= cost[i], target <= 5000

Code

DP

這題和 Coin Change II 類似，是 unbounded knapsack 的題目。DP 關係式為dp[i][j] = dp[i - 1][j] + (j >= cost[i - 1] ? dp[i][j - cost[i - 1]] : 0);

只是這題是 string，要注意 string 的相關操作。

另外，注意 base case 的設定，設為 ""，而 DP table 的初始值，設為 "0"，且在 for loop 中使用if(dp[i][j - cost[i - 1]] == "0") continue;，因為這題的 dp[i][j] 代表使用 index 0 ~ i 的 digit 組成 cost 剛剛好為 j 的 largest number。

Time Complexity: $O(NT)$, Space Complexity: $O(NT)$

class Solution {
    inline bool isGreaterThan(string &a, string &b) {
        return a.size() != b.size() ? a.size() > b.size() : a > b;
    }
public:
    string largestNumber(vector<int>& cost, int target) {
        int n = cost.size();
        vector<vector<string>> dp(n + 1, vector<string>(target + 1, "0")); // "0" means impossible
 
        // base case
        for(int i = 0; i < n + 1; i++) {
            dp[i][0] = "";
        }
 
        for(int i = 1; i < n + 1; i++) {
            for(int j = 1; j < target + 1; j++) {
                dp[i][j] = dp[i - 1][j];
                if(j >= cost[i - 1]) {
                    if(dp[i][j - cost[i - 1]] == "0") continue;
                    auto s = string(1, '0' + i) + dp[i][j - cost[i - 1]];
                    if (isGreaterThan(s, dp[i][j])) dp[i][j] = s;
                }
            }
        }
 
        return dp[n][target];
    }
};

DP - Optimized

Time Complexity: $O(NT)$, Space Complexity: $O(T)$

觀察 dp[i][j] 只和dp[i - 1][j] 和 dp[i][j - cost[i - 1]] 有關係，因此可以只用 1D dp table 。

class Solution {
    inline bool isGreaterThan(string &a, string &b) {
        return a.size() != b.size() ? a.size() > b.size() : a > b;
    }
public:
    string largestNumber(vector<int>& cost, int target) {
        int n = cost.size();
        vector<string> dp(target + 1, "0"); // "0" means impossible
 
        // base case
        dp[0] = "";
 
        for(int i = 1; i < n + 1; i++) {
            for(int j = 1; j < target + 1; j++) {
                if(j >= cost[i - 1]) {
                    if(dp[j - cost[i - 1]] == "0") continue;
                    auto s = string(1, '0' + i) + dp[j - cost[i - 1]];
                    if (isGreaterThan(s, dp[j])) dp[j] = s;
                }
            }
        }
 
        return dp[target];
    }
};

Source

• Form Largest Integer With Digits That Add up to Target - LeetCode