onenoughtone

Solution Approach

Solution Overview

There are 3 main approaches to solve this problem:

Dynamic Programming Approach - Complex approach
Greedy Approach - Complex approach
Recursive Approach with Memoization - Complex approach

Approach 1: Dynamic Programming Approach

The key insight for solving this problem is to use dynamic programming to determine if it's possible to form an integer with the given budget, and then construct the largest possible integer by prioritizing higher digits.

We'll define a DP array where dp[i] represents the maximum number of digits we can print with a budget of i. If dp[i] is -1, it means it's impossible to form an integer with a budget of i.

Once we've filled the DP array, we'll construct the largest possible integer by greedily selecting the highest possible digit at each step, starting from the most significant digit.

Algorithm:

Initialize a DP array of size target + 1 with -1, and set dp[0] = 0.
For each budget i from 1 to target:
a. For each digit j from 1 to 9:
i. If i >= cost[j-1] and dp[i - cost[j-1]] != -1, update dp[i] = max(dp[i], dp[i - cost[j-1]] + 1).
If dp[target] is -1, return an empty string (it's impossible to form any integer).
Construct the largest possible integer by greedily selecting the highest possible digit at each step:
a. Start with an empty result string and remaining budget = target.
b. For each position from 1 to dp[target]:
i. For each digit j from 9 down to 1:
1. If remaining budget >= cost[j-1] and dp[remaining budget - cost[j-1]] == dp[target] - position, append digit j to the result and update remaining budget -= cost[j-1].
2. Break the inner loop once a digit is selected.
Return the result string.

Time Complexity:

O(target)

Where target is the given budget. We fill a DP array of size target + 1, and for each position, we consider 9 possible digits, which is a constant factor.

Space Complexity:

O(target)

We use a DP array of size target + 1 to store the maximum number of digits we can print with each budget.

Approach 2: Greedy Approach

Another approach is to use a greedy algorithm to construct the largest possible integer. The idea is to prioritize higher digits and include as many of them as possible within the given budget.

However, this approach is more complex because we need to ensure that the resulting integer doesn't have leading zeros and that we maximize the number of digits.

Algorithm:

Find the digit with the minimum cost (let's call it minDigit and its cost minCost).
If target < minCost, return an empty string (it's impossible to form any integer).
Initialize an empty result string.
For each position from the most significant digit to the least significant digit:
a. For each digit j from 9 down to 1 (or 0 if it's not the most significant digit):
i. If target >= cost[j-1] and we can fill the remaining positions with at least minDigit (i.e., target - cost[j-1] >= minCost * (remainingPositions - 1)), append digit j to the result and update target -= cost[j-1].
ii. Break the inner loop once a digit is selected.
Return the result string.

Time Complexity:

O(target / minCost)

Where target is the given budget and minCost is the minimum cost among all digits. In the worst case, we can print target / minCost digits, and for each position, we consider at most 10 possible digits.

Space Complexity:

O(target / minCost)

We need to store the result string, which can have at most target / minCost digits.

Approach 3: Recursive Approach with Memoization

We can also solve this problem using recursion with memoization. The idea is to define a recursive function that returns the maximum number of digits we can print with a given budget, and then use this function to construct the largest possible integer.

This approach is similar to the dynamic programming approach but uses recursion instead of iteration.

Algorithm:

Define a recursive function maxDigits(budget) that returns the maximum number of digits we can print with the given budget:
a. If budget < 0, return -1 (invalid budget).
b. If budget == 0, return 0 (no more budget left).
c. If the result for budget is already memoized, return it.
d. Initialize result = -1.
e. For each digit j from 1 to 9:
i. Let subResult = maxDigits(budget - cost[j-1]).
ii. If subResult != -1, update result = max(result, subResult + 1).
f. Memoize and return result.
If maxDigits(target) is -1, return an empty string (it's impossible to form any integer).
Construct the largest possible integer using the maxDigits function:
a. Start with an empty result string and remaining budget = target.
b. For each position from 1 to maxDigits(target):
i. For each digit j from 9 down to 1 (or 0 if it's not the most significant digit):
1. If maxDigits(remaining budget - cost[j-1]) == maxDigits(target) - position, append digit j to the result and update remaining budget -= cost[j-1].
2. Break the inner loop once a digit is selected.
Return the result string.

Time Complexity:

O(target)

Where target is the given budget. We compute maxDigits(budget) for each budget from 1 to target, and for each budget, we consider 9 possible digits, which is a constant factor.

Space Complexity:

O(target)

We use memoization to store the results of maxDigits(budget) for each budget from 1 to target, which requires O(target) space.

Pseudocode

solution.pseudo

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function largestNumber(cost, target):
    // Initialize DP array
    dp = array of size target + 1, initialized with -1
    dp[0] = 0
 
    // Fill DP array
    for i from 1 to target:
        for j from 1 to 9:
            if i >= cost[j-1] and dp[i - cost[j-1]] != -1:
                dp[i] = max(dp[i], dp[i - cost[j-1]] + 1)
 
    // Check if it's possible to form any integer
    if dp[target] == -1:
        return ""
 
    // Construct the largest possible integer
    result = ""
    remaining = target
 
    for pos from 1 to dp[target]:
        for digit from 9 down to 1:
            if remaining >= cost[digit-1] and dp[remaining - cost[digit-1]] == dp[target] - pos:
                result += digit
                remaining -= cost[digit-1]
                break
 
    return result

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Learning Objective

Understand different approaches to solve the largest integer former problem and analyze their efficiency.