Find longest substring with no repeating characters
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Problem Statement
Given a string, find the length of the longest substring which has no repeating characters.
Example 1: Input: String=“aabccbb” Output: 3 Explanation: The longest substring without any repeating characters is “abc”.
Example 2: Input: String=“abbbb” Output: 2 Explanation: The longest substring without any repeating characters is “ab”.
Example 3: Input: String=“abccde” Output: 3 Explanation: Longest substrings without any repeating characters are “abc” & “cde”. Solution This problem follows the Sliding Window pattern and we can use a similar dynamic sliding window strategy as discussed in Longest Substring with K Distinct Characters. We can use a Map to remember the last index of each character we have processed. Whenever we get a repeating character we will shrink our sliding window to ensure that we always have distinct characters in the sliding window. We will find max window size in each pass and once the whole string is processed, the max window size is the total highest possible window size. Javascript code below:-
<script>
function non_repeat_substring(str) {
let windowStart = 0,
maxLength = 0,
charIndexMap = {};
// try to extend the range [windowStart, windowEnd]
for (let windowEnd = 0; windowEnd < str.length; windowEnd++) {
if (typeof charIndexMap[str[windowEnd]] !== 'undefined') {
windowStart = Math.max(windowStart, charIndexMap[str[windowEnd]]+1);
}
charIndexMap[str[windowEnd]] = windowEnd;
maxLength = Math.max(maxLength, windowEnd - windowStart + 1);
}
return maxLength;
}
document.write(`Length of the longest substring: ${non_repeat_substring('aabccbb')}<br/>`);
document.write(`Length of the longest substring: ${non_repeat_substring('abbbb')}<br/>`);
document.write(`Length of the longest substring: ${non_repeat_substring('abccde')}<br/>`);
</script>
Time Complexity
The time complexity of the above algorithm will be O(N) where ‘N’ is the number of characters in the input string.
Space Complexity
The space complexity of the algorithm will be O(K) where K is the number of distinct characters in the input string. This also means K<=N, because in the worst case, the whole string might not have any repeating character so the entire string will be added to the Map. Having said that, since we can expect a fixed set of characters in the input string (e.g., 26 for English letters), we can say that the algorithm runs in fixed space O(1); in this case, we can use a fixed-size array instead of the Map.
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