What is the brute force solution for Form Largest Integer With Digits That Add up to Target?

The brute force approach for Form Largest Integer With Digits That Add up to Target is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries to generate all possible combinations of digits that can be painted within the target cost. It checks each combination to see if it meets the cost requirement and keeps track of the largest number formed. The optimal Optimal Solution approach improves this to O(n * m).

What algorithm or data structure is used to solve Form Largest Integer With Digits That Add up to Target?

Form Largest Integer With Digits That Add up to Target uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Greedy Algorithms.

What are the interview tips for Form Largest Integer With Digits That Add up to Target?

Explain your thought process clearly as you write code. Consider edge cases, such as when the target is too small to paint any digit. Practice building the DP table step-by-step to reinforce understanding.

What are common mistakes when solving Form Largest Integer With Digits That Add up to Target?

Not initializing the DP array correctly. Failing to check if the previous state is empty before forming a new number.

#1449

Form Largest Integer With Digits That Add up to Target

Hard

ArrayDynamic ProgrammingDynamic ProgrammingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m)
Space	O(1)	O(n)

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Intuition

Time O(n * m)Space O(n)

Using dynamic programming, we can build a table that tracks the maximum number that can be formed for each possible cost up to the target. This allows us to efficiently find the largest number without generating all combinations.

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Algorithm

4 steps

1Step 1: Create a DP array where dp[i] stores the maximum number string that can be formed with cost i.
2Step 2: Initialize dp[0] to an empty string since no cost means no digits.
3Step 3: For each digit from 1 to 9, update the DP array by checking if the current cost can be formed by adding this digit.
4Step 4: After filling the DP array, return dp[target] or '0' if it remains empty.

solution.py11 lines

1# Full working Python code
2
3def largestInteger(cost, target):
4    dp = [''] * (target + 1)
5    dp[0] = ''
6    for digit in range(1, 10):
7        for t in range(cost[digit - 1], target + 1):
8            if dp[t - cost[digit - 1]] != '':
9                new_number = dp[t - cost[digit - 1]] + str(digit)
10                dp[t] = max(dp[t], new_number)
11    return dp[target] if dp[target] else '0'

ℹ

Complexity note: The complexity is O(n * m) where n is the target and m is the number of digits (which is constant at 9). This is efficient because we only iterate through the target cost and update the DP array.

1Dynamic programming allows us to build solutions incrementally.
2The order of digits matters for forming the largest number.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.