How do you solve DI String Match on LeetCode?

DI String Match (LeetCode #942) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The 'I' and 'D' characters dictate the order of the numbers directly.

What is the brute force solution for DI String Match?

The brute force approach for DI String Match is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible permutations of the numbers from 0 to n and checking which one matches the given string pattern. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve DI String Match?

DI String Match uses the following concepts: Array, Two Pointers, String, Greedy. The recommended patterns to study are: Two Pointers, Greedy.

What are the interview tips for DI String Match?

Think about how to construct the result based on the input constraints. Practice recognizing patterns in problems, like using two pointers or greedy strategies. Always clarify the problem constraints and requirements before diving into coding.

What are common mistakes when solving DI String Match?

Not understanding how to use the two pointers effectively. Overcomplicating the problem by trying to generate permutations instead of constructing the result directly.

#942

DI String Match

Easy

ArrayTwo PointersStringGreedyTwo PointersGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal approach uses a greedy strategy to construct the permutation directly based on the 'I' and 'D' characters in the string. We can maintain two pointers to fill in the permutation efficiently.

⚙️

Algorithm

3 steps

1Step 1: Initialize an array perm of size n + 1 with values from 0 to n.
2Step 2: Use two pointers: one starting from the beginning (left) and one from the end (right) of the array.
3Step 3: Iterate through the string s: if s[i] is 'I', assign perm[left] to the result and increment left; if 'D', assign perm[right] to the result and decrement right.

solution.py14 lines

1def diStringMatch(s):
2    n = len(s)
3    perm = list(range(n + 1))
4    left, right = 0, n
5    result = []
6    for char in s:
7        if char == 'I':
8            result.append(perm[left])
9            left += 1
10        else:
11            result.append(perm[right])
12            right -= 1
13    result.append(perm[left])  # Add the last element
14    return result

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the string s and fill the result array in linear time. The space complexity is O(n) due to the result array.

1The 'I' and 'D' characters dictate the order of the numbers directly.
2Using two pointers allows us to fill the permutation efficiently without generating all possibilities.

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