Max Consecutive Ones

easy array

Problem

Given a binary array nums, return the maximum number of consecutive 1s in the array.

Inputnums = [1, 1, 0, 1, 1, 1]

Output3

The longest run of 1s is at the tail.

nums (binary, comma-separated)

def find_max_consecutive_ones(nums):
    best = run = 0
    for x in nums:
        run = run + 1 if x == 1 else 0
        if run > best: best = run
    return best

function findMaxConsecutiveOnes(nums) {
  let best = 0, run = 0;
  for (const x of nums) {
    run = x === 1 ? run + 1 : 0;
    if (run > best) best = run;
  }
  return best;
}

class Solution {
    public int findMaxConsecutiveOnes(int[] nums) {
        int best = 0, run = 0;
        for (int x : nums) {
            run = x == 1 ? run + 1 : 0;
            if (run > best) best = run;
        }
        return best;
    }
}

int findMaxConsecutiveOnes(vector<int>& nums) {
    int best = 0, run = 0;
    for (int x : nums) {
        run = x == 1 ? run + 1 : 0;
        if (run > best) best = run;
    }
    return best;
}

Explanation

We want the longest unbroken stretch of 1s. The simple trick is to sweep through once with a running counter that grows on every 1 and snaps back to zero on every 0.

We keep run (length of the current streak of 1s) and best (longest streak seen). When the value is 1, do run = run + 1; otherwise reset run = 0.

After each step we update best if the current run is larger. Because the reset happens exactly where a 0 breaks the streak, run always reflects the length ending at the current position.

This single pass is enough — we never need to revisit earlier elements.

Example: [1, 1, 0, 1, 1, 1]. run reaches 2 over the first pair, resets at the 0, then climbs to 3 over the final three 1s, so best = 3.

Time: O(n) Space: O(1)