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longest_substring_k_distinct_test.go
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longest_substring_k_distinct_test.go
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/*
Problem:
- Given a string, find the length of the longest substring in it with no more
than k distinct characters.
Example:
- Input: string="araaci", k=1
Output: 2
Explanation: Longest substring with no more than 1 distinct characters is "aa".
- Input: string="araaci", k=2
Output: 4
Explanation: Longest substring with no more than 2 distinct characters is "araa".
- Input: string="araaci", k=3
Output: 5
Explanation: Longest substring with no more than 3 distinct characters is "araac".
Approach:
- Use a hashmap to remember the frequency of each character we have seen.
- Insert characters until we have k distinct characters in the map to be consider a
window.
- Shrink the window until there is no more k distinct characters in the map and keep
updating the maximum window length at each step.
Cost:
- O(n) time, O(k) space where k is the number of characters in the map.
*/
package gtci
import (
"testing"
"github.com/hoanhan101/algo/common"
)
func TestLongestSubstringKDistinct(t *testing.T) {
tests := []struct {
in1 string
in2 int
expected int
}{
{"", 0, 0},
{"", 1, 0},
{"a", 0, 0},
{"a", 1, 1},
{"aa", 1, 2},
{"aa", 2, 2},
{"ab", 1, 1},
{"ab", 2, 2},
{"araaci", 1, 2},
{"araaci", 2, 4},
{"araaci", 3, 5},
}
for _, tt := range tests {
common.Equal(
t,
tt.expected,
longestSubstringKDistinct(tt.in1, tt.in2),
)
}
}
func longestSubstringKDistinct(s string, k int) int {
maxLength, start := 0, 0
// char keeps track of characters' frequencies.
char := map[string]int{}
for end := range s {
// insert characters until we have k distinct characters.
endChar := string(s[end])
if _, ok := char[endChar]; !ok {
char[endChar] = 0
}
char[endChar]++
// shrink the window until there is no more than k distinct characters.
for len(char) > k {
startChar := string(s[start])
// decrement the frequency of the one going out of the window and
// remove if its frequency is zero.
char[startChar]--
if char[startChar] == 0 {
delete(char, startChar)
}
// increase the start index to move the window ahead by one element.
start++
}
// update the maximum length at each step.
maxLength = common.Max(maxLength, end-start+1)
}
return maxLength
}