Given two strings s1 and s2, check if s2 contains a permutation of s1. Return true if s2 contains any permutation of s1, otherwise return false. Pseudocode: count1 = frequency array of s1window = frequency array of the first len(s1) characters of s2for i from len(s1) to len(s2): if window matches count1: return true slide window by updating frequenciesif window matches count1: return truereturn false Time Complexity: O(n * m), Space Complexity: O(1) Time Complexity: O(n), Space Complexity: O(m) Time Complexity: O(n), Space Complexity: O(1) Time Complexity: O(n^2), Space Complexity: O(n)
Given two strings s1 and s2, check if s2 contains a permutation of s1. Return true if s2 contains any permutation of s1, otherwise return false. Pseudocode: count1 = frequency array of s1window = frequency array of the first len(s1) characters of s2for i from len(s1) to len(s2): if window matches count1: return true slide window by updating frequenciesif window matches count1: return truereturn false Time Complexity: O(n * m), Space Complexity: O(1) Time Complexity: O(n), Space Complexity: O(m) Time Complexity: O(n), Space Complexity: O(1) Time Complexity: O(n^2), Space Complexity: O(n)
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Question
Given two strings s1 and s2, check if s2 contains a permutation of s1. Return true if s2 contains any permutation of s1, otherwise return false.
Pseudocode:
count1 = frequency array of s1window = frequency array of the first len(s1) characters of s2
for i from len(s1) to len(s2):
if window matches count1:
return true
slide window by updating frequencies
if window matches count1:
return true
return false
Time Complexity: O(n * m), Space Complexity: O(1)
Time Complexity: O(n), Space Complexity: O(m)
Time Complexity: O(n), Space Complexity: O(1)
Time Complexity: O(n^2), Space Complexity: O(n)
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