1449. 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

Solutions

class Solution {
    public String largestNumber(int[] cost, int target) {
        final int inf = 1 << 30;
        int[][] f = new int[10][target + 1];
        int[][] g = new int[10][target + 1];
        for (var e : f) {
            Arrays.fill(e, -inf);
        }
        f[0][0] = 0;
        for (int i = 1; i <= 9; ++i) {
            int c = cost[i - 1];
            for (int j = 0; j <= target; ++j) {
                if (j < c || f[i][j - c] + 1 < f[i - 1][j]) {
                    f[i][j] = f[i - 1][j];
                    g[i][j] = j;
                } else {
                    f[i][j] = f[i][j - c] + 1;
                    g[i][j] = j - c;
                }
            }
        }
        if (f[9][target] < 0) {
            return "0";
        }
        StringBuilder sb = new StringBuilder();
        for (int i = 9, j = target; i > 0;) {
            if (j == g[i][j]) {
                --i;
            } else {
                sb.append(i);
                j = g[i][j];
            }
        }
        return sb.toString();
    }
}

class Solution {
public:
    string largestNumber(vector<int>& cost, int target) {
        const int inf = 1 << 30;
        vector<vector<int>> f(10, vector<int>(target + 1, -inf));
        vector<vector<int>> g(10, vector<int>(target + 1));
        f[0][0] = 0;
        for (int i = 1; i <= 9; ++i) {
            int c = cost[i - 1];
            for (int j = 0; j <= target; ++j) {
                if (j < c || f[i][j - c] + 1 < f[i - 1][j]) {
                    f[i][j] = f[i - 1][j];
                    g[i][j] = j;
                } else {
                    f[i][j] = f[i][j - c] + 1;
                    g[i][j] = j - c;
                }
            }
        }
        if (f[9][target] < 0) {
            return "0";
        }
        string ans;
        for (int i = 9, j = target; i;) {
            if (g[i][j] == j) {
                --i;
            } else {
                ans += '0' + i;
                j = g[i][j];
            }
        }
        return ans;
    }
};

class Solution:
    def largestNumber(self, cost: List[int], target: int) -> str:
        f = [[-inf] * (target + 1) for _ in range(10)]
        f[0][0] = 0
        g = [[0] * (target + 1) for _ in range(10)]
        for i, c in enumerate(cost, 1):
            for j in range(target + 1):
                if j < c or f[i][j - c] + 1 < f[i - 1][j]:
                    f[i][j] = f[i - 1][j]
                    g[i][j] = j
                else:
                    f[i][j] = f[i][j - c] + 1
                    g[i][j] = j - c
        if f[9][target] < 0:
            return "0"
        ans = []
        i, j = 9, target
        while i:
            if j == g[i][j]:
                i -= 1
            else:
                ans.append(str(i))
                j = g[i][j]
        return "".join(ans)

func largestNumber(cost []int, target int) string {
	const inf = 1 << 30
	f := make([][]int, 10)
	g := make([][]int, 10)
	for i := range f {
		f[i] = make([]int, target+1)
		g[i] = make([]int, target+1)
		for j := range f[i] {
			f[i][j] = -inf
		}
	}
	f[0][0] = 0
	for i := 1; i <= 9; i++ {
		c := cost[i-1]
		for j := 0; j <= target; j++ {
			if j < c || f[i][j-c]+1 < f[i-1][j] {
				f[i][j] = f[i-1][j]
				g[i][j] = j
			} else {
				f[i][j] = f[i][j-c] + 1
				g[i][j] = j - c
			}
		}
	}
	if f[9][target] < 0 {
		return "0"
	}
	ans := []byte{}
	for i, j := 9, target; i > 0; {
		if g[i][j] == j {
			i--
		} else {
			ans = append(ans, '0'+byte(i))
			j = g[i][j]
		}
	}
	return string(ans)
}

function largestNumber(cost: number[], target: number): string {
    const inf = 1 << 30;
    const f: number[][] = Array(10)
        .fill(0)
        .map(() => Array(target + 1).fill(-inf));
    const g: number[][] = Array(10)
        .fill(0)
        .map(() => Array(target + 1).fill(0));
    f[0][0] = 0;
    for (let i = 1; i <= 9; ++i) {
        const c = cost[i - 1];
        for (let j = 0; j <= target; ++j) {
            if (j < c || f[i][j - c] + 1 < f[i - 1][j]) {
                f[i][j] = f[i - 1][j];
                g[i][j] = j;
            } else {
                f[i][j] = f[i][j - c] + 1;
                g[i][j] = j - c;
            }
        }
    }
    if (f[9][target] < 0) {
        return '0';
    }
    const ans: number[] = [];
    for (let i = 9, j = target; i; ) {
        if (g[i][j] === j) {
            --i;
        } else {
            ans.push(i);
            j = g[i][j];
        }
    }
    return ans.join('');
}

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

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

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

Description

Solutions

All Problems

All Solutions