1869. Longer Contiguous Segments of Ones than Zeros

Method 1

Given a binary string s, return true if the longest contiguous segment of 1‘s is strictly longer than the longest contiguous segment of 0‘s in s, or return false otherwise.
- For example, in s = "110100010" the longest continuous segment of 1s has length 2, and the longest continuous segment of 0s has length 3.
Note that if there are no 0’s, then the longest continuous segment of 0’s is considered to have a length 0. The same applies if there is no 1’s.

Example 1

Input: s = "1101"
Output: true
Explanation:
The longest contiguous segment of 1s has length 2: "1101"
The longest contiguous segment of 0s has length 1: "1101"
The segment of 1s is longer, so return true.

Example 2

Input: s = "111000"
Output: false
Explanation:
The longest contiguous segment of 1s has length 3: "111000"
The longest contiguous segment of 0s has length 3: "111000"
The segment of 1s is not longer, so return false.

Example 3

Input: s = "110100010"
Output: false
Explanation:
The longest contiguous segment of 1s has length 2: "110100010"
The longest contiguous segment of 0s has length 3: "110100010"
The segment of 1s is not longer, so return false.

Method 1

【O(n) time | O(1) space】

package Leetcode.GroupedLoop;

/**
 * @author zhengxingxing
 * @date 2025/04/09
 */
public class LongerContiguousSegmentsOfOnesThanZeros {

    public static boolean checkZeroOnes(String s) {
        // Get the length of input string
        int n = s.length();

        // Variables to track the maximum consecutive counts of 1's and 0's
        int maxOnes = 0;
        int maxZeros = 0;

        // Index for traversing the string
        int i = 0;

        // Process the string using grouped loop pattern
        while (i < n) {
            // Mark the start position of current group
            int start = i;

            // Get the character that defines current group (either '0' or '1')
            char currentChar = s.charAt(start);

            // Continue while we see the same character
            while (i < n && s.charAt(i) == currentChar) {
                i++;
            }

            // Calculate the length of current continuous group
            int groupLength = i - start;

            // Update the appropriate maximum count based on current character
            if (currentChar == '1') {
                maxOnes = Math.max(maxOnes, groupLength);
            } else {
                maxZeros = Math.max(maxZeros, groupLength);
            }
        }

        // Return true if longest segment of 1's is strictly longer than 0's
        return maxOnes > maxZeros;
    }
    
    public static void main(String[] args) {
        // Test Case 1: Basic case with mixed 1's and 0's
        String test1 = "1101";
        System.out.println("Test 1: " + test1);
        System.out.println("Expected result: true");
        System.out.println("Actual result: " + checkZeroOnes(test1));
        System.out.println();

        // Test Case 2: Equal length segments
        String test2 = "11000";
        System.out.println("Test 2: " + test2);
        System.out.println("Expected result: false");
        System.out.println("Actual result: " + checkZeroOnes(test2));
        System.out.println();

        // Test Case 3: Single character
        String test3 = "1";
        System.out.println("Test 3: " + test3);
        System.out.println("Expected result: true");
        System.out.println("Actual result: " + checkZeroOnes(test3));
        System.out.println();

        // Test Case 4: Alternating pattern
        String test4 = "101010";
        System.out.println("Test 4: " + test4);
        System.out.println("Expected result: false");
        System.out.println("Actual result: " + checkZeroOnes(test4));
        System.out.println();

        // Test Case 5: Long continuous sequence
        String test5 = "111000111111";
        System.out.println("Test 5: " + test5);
        System.out.println("Expected result: true");
        System.out.println("Actual result: " + checkZeroOnes(test5));
        System.out.println();
    }
}

Method 1

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