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This approach involves iterating through the array and counting sequences of 1s. If a 0 is encountered, the count is reset to 0. We keep track of the maximum count during the iteration.
Time Complexity: O(n), where n is the number of elements in the array, as we make a single pass.
Space Complexity: O(1) since no extra space proportional to input size is used.
1function findMaxConsecutiveOnes(nums) {
2 let maxCount = 0;
3 let currentCount = 0;
4 for(let num of nums) {
5 if(num === 1) {
6 currentCount++;
7 maxCount = Math.max(maxCount, currentCount);
8 } else {
9 currentCount = 0;
10 }
11 }
12 return maxCount;
13}
14
15let nums = [1, 1, 0, 1, 1, 1];
16console.log(`The maximum number of consecutive 1s is: ${findMaxConsecutiveOnes(nums)}`);This JavaScript function uses a for...of loop to iterate through the array. It keeps track of the current and maximum number of consecutive 1s.
In this version, two pointers left and right maintain the window of consecutive 1s. When a 0 is found, left is moved to right + 1, effectively 'skipping' the segments with 0s, adjusting the window accordingly.