3632. Button With Longest Push Time¶

3632. Button with Longest Push Time

Easy

You are given a 2D array events which represents a sequence of events where a child pushes a series of buttons on a keyboard.

Each events[i] = [indexi, timei] indicates that the button at index indexi was pressed at time timei.

• The array is sorted in increasing order of time.
• The time taken to press a button is the difference in time between consecutive button presses. The time for the first button is simply the time at which it was pressed.

Return the index of the button that took the longest time to push. If multiple buttons have the same longest time, return the button with the smallest index.

Example 1:

Example 2:

Constraints:

• 1 <= events.length <= 1000
• events[i] == [indexi, timei]
• 1 <= indexi, timei <= 105
• The input is generated such that events is sorted in increasing order of timei.

Solution¶

class Solution {
    public int buttonWithLongestTime(int[][] events) {
        int n = events.length;
        int maxi_diff = 0, ans = n;
        for (int i = 0; i < n - 1; i++) {
            int current = events[i][1];
            int next = events[i + 1][1];
            if (next - current > maxi_diff) {
                maxi_diff = next - current;
                ans = events[i + 1][0];
                ans = Math.min(ans, events[i + 1][0]);
            }
            else if (next - current == maxi_diff) ans = Math.min(ans, events[i + 1][0]);
        }
        int current = 0;
        int next = events[0][1];
        if (next - current > maxi_diff) {
            maxi_diff = next - current;
            ans = events[0][0];
        } 
        else if (next - current == maxi_diff) ans = Math.min(ans, events[0][0]);
        return ans;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Explanation¶

[Add detailed explanation here]