algoadvance

You are given a string s that consists only of the characters 'I' (increase) and 'D' (decrease). Assume that s has a length n. You need to find a permutation perm of the integers from 0 to n such that for all i:

• If s[i] == 'I', then perm[i] < perm[i + 1]
• If s[i] == 'D', then perm[i] > perm[i + 1]

Clarifying Questions

1. Input Constraints
   • Is the input string guaranteed to have only characters ‘I’ and ‘D’?
     • Yes, the problem guarantees this constraint.
   • What is the range for the length n?
     • The length can be up to 10,000, as per typical LeetCode problem constraints.
2. Output Expectations
   • Should the output be any valid permutation that satisfies the condition?
     • Yes, any valid permutation meeting the criteria is acceptable.

Strategy

To solve this problem, we can use a two-pointer technique. Start with the smallest and largest possible numbers, and place them based on the characters in the string:

• Initialize two pointers: low to 0 and high to n (where n is the length of the string s).
• Traverse through the string s and construct the permutation:
  • If the current character is 'I', set the current position in the permutation to low and increment low.
  • If the current character is 'D', set the current position in the permutation to high and decrement high.
• The last element to be placed after the loop will be the remaining value of low (which should be equal to high at that point).

This approach ensures that all conditions are met with a time complexity of O(n).

Code

def diStringMatch(s: str):
    n = len(s)
    low, high = 0, n
    perm = []
    
    for char in s:
        if char == 'I':
            perm.append(low)
            low += 1
        else:
            perm.append(high)
            high -= 1
    
    # Append the last remaining number which should be same for both low and high
    perm.append(low)  # low and high are the same at this point
    
    return perm

# Example Usage
s = "IDID"
print(diStringMatch(s))  # Output should be a valid permutation like [0, 4, 1, 3, 2]

Time Complexity

• Time Complexity: O(n), where n is the length of the string s. We make a single pass through the string to determine the permutation.
• Space Complexity: O(n), where n is the length of the string s, due to the storage requirement of the output list perm.

Try our interview co-pilot at AlgoAdvance.com

• Got blindsided by a question you didn’t expect?

• Spend too much time studying?

• Or simply don’t have the time to go over all 3000 questions?