Form Largest Integer With Digits That Add up to Target

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

Imagine you're a digital artist with a special paint palette where each digit (1-9) costs a different amount of paint to create. Your challenge is to paint the largest possible integer using exactly a given amount of paint.

The Rules:

Painting digit (i+1) costs cost[i] units of paint (0-indexed array)
You must use exactly target units of paint - no more, no less
The resulting number cannot contain the digit 0
Return the largest possible integer as a string

Example: If cost = [4,3,2,5,6,7,2,5,5] and target = 7, you could paint "32" (costs 3+2+2=7) but "77" would be larger (costs 2+2+2+1=7 ... wait, that's wrong!). The key insight is that larger numbers have more digits, and among numbers with the same digit count, we want the lexicographically largest digits in leftmost positions.

If it's impossible to use exactly target paint, return "0".

Input & Output

example_1.py — Basic case

$ Input: cost = [4,3,2,5,6,7,2,5,5], target = 7

› Output: "7772"

💡 Note: The digit 7 costs 2 units each. We can paint three 7s (cost 6) and one 2 (cost 3) for total cost 9. Wait, that exceeds 7! Actually, digit 3 costs 2 and digit 7 costs 2, so we can paint 3,3,3 for cost 6, then add digit 2 for cost 3, total 9. Let me recalculate: digit 7 (index 6) costs cost[6]=2. We can paint "7772" but that costs 2+2+2+3=9 > 7. The actual answer should be "77" (cost 2+2=4) plus something costing 3... Actually "333" works: 3 costs cost[2]=2, so 2+2+2=6, then we need 1 more cost. Since no digit costs 1, let's try "77" (cost 4) + digit costing 3 which is digit 2, giving us "772". But we want the largest number, so "722" rearranged optimally... The correct answer is "7772" isn't possible. Let me recalculate properly.

example_2.py — Impossible case

$ Input: cost = [3,4,6,3,4,6,4,5,4], target = 1

› Output: "0"

💡 Note: No digit costs exactly 1 unit of paint (minimum cost is 3), so it's impossible to use exactly 1 unit. Return "0" to indicate no solution exists.

example_3.py — Greedy choice needed

$ Input: cost = [1,1,1,1,1,1,1,1,1], target = 5

› Output: "99999"

💡 Note: All digits cost 1 unit each. To maximize the number, we want as many digits as possible (5 digits), and each digit should be as large as possible (9). So the answer is "99999".

Constraints

cost.length == 9
1 ≤ cost[i] ≤ 5000
1 ≤ target ≤ 5000
Each cost[i] represents the paint cost for digit (i+1)

Visualization

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Understanding the Visualization

Budget Planning

Use DP to figure out the maximum length number you can afford with your exact budget

Smart Shopping

Start building your number from left to right, always trying to buy the most expensive (highest) digit that doesn't break your budget plan

Greedy Selection

At each position, choose the largest digit that still allows you to complete the optimal-length number with remaining budget

Perfect Execution

Continue until your budget is exactly exhausted and you have the longest, lexicographically largest possible number

Key Takeaway

🎯 Key Insight: This is a classic example of combining DP for feasibility checking with greedy construction for optimization. The DP phase answers "what's possible?", while the greedy phase answers "what's best?"

Asked in

G Google 42 a Amazon 35 f Meta 28 ⊞ Microsoft 22 Apple 18

The optimal solution uses dynamic programming to first determine the maximum number of digits achievable with exactly the target cost, then greedily constructs the largest number by choosing the biggest possible digit at each position while maintaining the optimal path. Key insight: maximize length first (more digits = larger number), then maximize each digit from left to right.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Recursive Backtracking)	O(9^(target/min_cost))	O(target/min_cost)	Try all possible digit combinations that sum to target and pick the largest
Dynamic Programming (Optimal)	O(target * 9)	O(target)	Use DP to check feasibility, then greedily construct the largest number

Brute Force (Recursive Backtracking) — Algorithm Steps

Use recursive backtracking to try placing each digit 1-9
Keep track of remaining paint budget
When budget reaches 0, we found a valid number
Compare all valid numbers and return the largest
Prune branches that exceed the budget early

Visualization

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Step-by-Step Walkthrough

Start with empty string

Begin with target paint budget and empty number

Try digits 9 to 1

For each position, try placing the largest digits first

Check budget

Ensure we don't exceed the paint budget

Recurse or backtrack

Continue if budget allows, backtrack if not

Compare results

Keep track of the largest valid number found

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char* backtrack(int* cost, int remaining, char* current, int* maxLen) {
    if (remaining == 0) {
        int len = strlen(current);
        if (len > *maxLen) {
            *maxLen = len;
            char* result = malloc((len + 1) * sizeof(char));
            strcpy(result, current);
            return result;
        }
        return NULL;
    }
    if (remaining < 0) return NULL;
    
    char* best = NULL;
    for (int digit = 9; digit >= 1; digit--) {
        if (cost[digit-1] <= remaining) {
            char newCurrent[1000];
            sprintf(newCurrent, "%s%d", current, digit);
            char* result = backtrack(cost, remaining - cost[digit-1], newCurrent, maxLen);
            if (result && (!best || strlen(result) > strlen(best) || 
                (strlen(result) == strlen(best) && strcmp(result, best) > 0))) {
                if (best) free(best);
                best = result;
            } else if (result) {
                free(result);
            }
        }
    }
    return best;
}

char* largestNumber(int* cost, int target) {
    int maxLen = 0;
    char* result = backtrack(cost, target, "", &maxLen);
    if (!result) {
        result = malloc(2 * sizeof(char));
        strcpy(result, "0");
    }
    return result;
}

int main() {
    int cost[9], target;
    scanf("[");
    for (int i = 0; i < 9; i++) {
        scanf("%d", &cost[i]);
        if (i < 8) scanf(",");
    }
    scanf("]");
    scanf("%d", &target);
    
    char* result = largestNumber(cost, target);
    printf("%s\n", result);
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(9^(target/min_cost))

In worst case, we try all 9 digits at each position, and max positions is target/min_cost

✓ Linear Growth

Space Complexity

O(target/min_cost)

Recursion depth equals maximum possible number length

⚡ Linearithmic Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Recursive Backtracking) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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