Given two strings needle and haystack, return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.
Example 1:
Input: haystack = "sadbutsad", needle = "sad"
Output: 0
Explanation: "sad" occurs at index 0 and 6.
The first occurrence is at index 0, so we return 0.
Example 2:
Input: haystack = "leetcode", needle = "leeto"
Output: -1
Explanation: "leeto" did not occur in "leetcode", so we return -1.
Constraints:
1 <= haystack.length, needle.length <= 104haystack and needle consist of only lowercase English characters.
KMP
Preprocess needle to generate the longest_border array
e.g.
a b c d a b e e a b f
0 0 0 0 1 2 0 0 1 2 0
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class Solution:
def strStr(self, haystack: str, needle: str) -> int:
m = len(needle)
n = len(haystack)
if n < m:
return -1
# == PREPROCESSING
# longest border array
longest_border = [0] * m
# Length of Longest Border for prefix before it.
prev = 0
# Iterating from index-1. longest_border[0] will always be 0
i = 1
while i < m:
if needle[i] == needle[prev]:
# Length of Longest Border Increased
prev += 1
longest_border[i] = prev
i += 1
else:
# Only empty border exist
if prev == 0:
longest_border[i] = 0
i += 1
# Try finding longest border for this i with reduced prev
else:
prev = longest_border[prev - 1]
# == SEARCHING
# Pointer for haystack
haystack_pointer = 0
# Pointer for needle.
# Also indicates number of characters matched in current window.
needle_pointer = 0
while haystack_pointer < n:
if haystack[haystack_pointer] == needle[needle_pointer]:
# Matched Increment Both
needle_pointer += 1
haystack_pointer += 1
# All characters matched
if needle_pointer == m:
# m characters behind last matching will be window start
return haystack_pointer - m
else:
if needle_pointer == 0:
# Zero Matched
haystack_pointer += 1
else:
# Optimally shift left needle_pointer.
# Don't change haystack_pointer
needle_pointer = longest_border[needle_pointer - 1]
return -1
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