Codility Test Synechron|Online real-time assessment|Synechron Java developers 2024

This Test Is for Java backend developer and for 4+ years of experience in same.
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[@abhishekpatel6020]

Nice question

[@dricmoybhatthacharjee8119]

Clean solution :)

[@HappyTechies]

class Solution {
public int solution(String S) {
int patches = 0;
int i = 0;
while (i < S.length()) {
if (S.charAt(i) == 'X') {
// Place a patch covering segments i, i+1, and i+2
patches++;
i += 3;
} else {
i++;
}
}
return patches;
}
}
Explanation:
Initialize patches to 0, which will count the number of patches used.
Initialize i to 0, which will be used to traverse the string.
While i is less than the length of the string S:
If the current character S.charAt(i) is 'X', it means we need to place a patch.
Increment the patches counter by 1 and move i by 3 positions to the right because the patch covers the current segment and the next two segments.
If the current character is not 'X', just move i by 1 position to the right.
This algorithm ensures that you cover all potholes with the minimum number of patches. The time complexity is O(N), where N is the length of the string S, making it efficient for the given constraints.