LeetCode 1449: Form Largest Integer With Digits That Add up to Target Solution

Master LeetCode problem 1449 (Form Largest Integer With Digits That Add up to Target), 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.

1449. Form Largest Integer With Digits That Add up to Target

Hard

Array Dynamic Programming

Problem Explanation

Explanation:

To solve this problem, we can use dynamic programming. We create a 1D array dp where dp[i] represents the maximum number we can form with a cost equal to i. We iterate over each digit cost and update the dp array accordingly. Finally, we build the result string by backtracking through the dp array.

Initialize dp array with size target + 1 and fill it with -1 (to indicate it's not possible to form a number with that cost).
Set dp[0] = 0 as the cost of forming number '0' is 0.
Iterate over each digit cost and update dp array.
Backtrack through the dp array to build the result string.

Time Complexity: O(target) where target is the given target value. Space Complexity: O(target) for the dp array.

Solution Code

class Solution {
    public String largestNumber(int[] cost, int target) {
        int[] dp = new int[target + 1];
        Arrays.fill(dp, -1);
        dp[0] = 0;
        
        for (int i = 1; i <= target; i++) {
            for (int j = 0; j < 9; j++) {
                if (i - cost[j] >= 0 && dp[i - cost[j]] >= 0) {
                    dp[i] = Math.max(dp[i], dp[i - cost[j]] + 1);
                }
            }
        }
        
        if (dp[target] < 0) {
            return "0";
        }
        
        StringBuilder sb = new StringBuilder();
        for (int i = 8; i >= 0; i--) {
            while (target - cost[i] >= 0 && dp[target] == dp[target - cost[i]] + 1) {
                sb.append(i + 1);
                target -= cost[i];
            }
        }
        
        return sb.toString();
    }
}

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