1784. Check if Binary String Has at Most One Segment of Ones
Explanation
To solve this problem, we can iterate through the binary string and check if there is more than one segment of ones. We can maintain a flag to track if we have seen a one and if we see another one, we can set the flag to false. If we encounter a zero after seeing a one, we can reset the flag. If the flag is false and we encounter another one, then there are more than one segment of ones.
- Initialize a boolean flag
foundOneto false. - Iterate through the binary string.
- If the current character is '1', check if
foundOneis already true. If true, return false as there are more than one segment of ones. Otherwise, setfoundOneto true. - If the current character is '0' and
foundOneis true, resetfoundOneto false. - If the loop finishes without any issues, return true as there is at most one segment of ones.
Time Complexity: O(n), where n is the length of the input string. Space Complexity: O(1)
class Solution {
public boolean checkOnesSegment(String s) {
boolean foundOne = false;
for (char c : s.toCharArray()) {
if (c == '1') {
if (foundOne) {
return false;
}
foundOne = true;
} else if (foundOne) {
foundOne = false;
}
}
return true;
}
}Code Editor (Testing phase)
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