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How to count the number of non-overlapping instances of a substring in a string in Python

1 Answer

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# Non‑overlapping occurrences are matches of a substring that do not reuse any of 
# the same characters. Once one match is counted, the next search must begin 
# after that match ends.
 
def count_non_overlapping(haystack: str, needle: str) -> int:
    """
    Count how many times 'needle' appears in 'haystack' without overlapping.
    The algorithm:
      • Use str.find() to locate the next occurrence.
      • Each time a match is found, move the search index forward
        by the full length of the matched substring.
      • This ensures no characters are reused between matches.
    """
    count = 0
    index = 0  # current search position in the main string
 
    # Continue searching until .find() returns -1 (meaning: no more matches)
    while True:
        # Find the next occurrence starting at the current index
        pos = haystack.find(needle, index)
 
        if pos == -1:
            # No more matches found
            break
 
        # We found a match, so increment the count
        count += 1
 
        # Move index forward by the length of the needle
        # This ensures the next search begins *after* the matched substring
        index = pos + len(needle)
 
    return count
 
 
# ---------------------------------------------------------------
s = 'go java phphp rust c pphpp c++ phpphp python php phphp'
substring = 'php'
 
# Count non-overlapping occurrences
result = count_non_overlapping(s, substring)
 
print("Non-overlapping occurrences:", result)


 
'''
run:
 
Non-overlapping occurrences: 6
 
'''

 



answered Aug 24, 2024 by avibootz
edited Jul 18 by avibootz

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