Button with Longest Push Time

• Create an auxiliary array durations of size N x 2 to store [duration, index].
• Initialize previousTime = 0.
• Iterate through the events array from i = 0 to N-1:
  • Calculate duration = events[i][1] - previousTime.
  • Store duration and events[i][0] in the durations array at index i.
  • Update previousTime = events[i][1].
• Initialize longestDuration = -1 and resultIndex = -1.
• Iterate through the durations array:
  • If the current duration is greater than longestDuration, update longestDuration and resultIndex.
  • If the current duration equals longestDuration, update resultIndex to the minimum of the current resultIndex and the current index.
• Return resultIndex.

We can use a list of pairs or a 2D array to store the calculated (duration, index) for each event.

We initialize a previousTime variable to 0. We loop through the input events array. For each event, we calculate the duration as currentTime - previousTime. We then add a new pair (duration, buttonIndex) to our auxiliary storage. After processing the event, we update previousTime to currentTime for the next iteration.

After the first loop populates our storage, we initialize maxDuration and resultIndex. We then loop through our stored (duration, index) pairs. In this loop, we compare each duration with maxDuration.

• If the current duration is greater than maxDuration, we update maxDuration and set resultIndex to the current index.
• If the current duration is equal to maxDuration, we update resultIndex only if the current index is smaller than the existing resultIndex, as per the tie-breaking rule.

Finally, we return the resultIndex.

1class Solution {2    public int longestPush(int[][] events) {3        if (events.length == 0) {4            return -1; // Should not happen based on constraints5        }6 7        // Use a 2D array to store [duration, index] for each event8        int[][] durations = new int[events.length][2];9        int previousTime = 0;10 11        // First pass: calculate and store all durations12        for (int i = 0; i < events.length; i++) {13            int index = events[i][0];14            int time = events[i][1];15            int duration = time - previousTime;16            durations[i][0] = duration;17            durations[i][1] = index;18            previousTime = time;19        }20 21        // Second pass: find the longest duration and corresponding index22        int longestDuration = -1;23        int resultIndex = -1;24 25        for (int i = 0; i < durations.length; i++) {26            int duration = durations[i][0];27            int index = durations[i][1];28            if (duration > longestDuration) {29                longestDuration = duration;30                resultIndex = index;31            } else if (duration == longestDuration) {32                if (index < resultIndex) {33                    resultIndex = index;34                }35            }36        }37 38        return resultIndex;39    }40}

• Initialize longestDuration = -1, resultIndex = -1, and previousTime = 0.
• Iterate through each event in the events array:
  • Let currentIndex = event[0] and currentTime = event[1].
  • Calculate duration = currentTime - previousTime.
  • If duration > longestDuration:
    • Set longestDuration = duration.
    • Set resultIndex = currentIndex.
  • Else if duration == longestDuration:
    • Set resultIndex = min(resultIndex, currentIndex).
  • Update previousTime = currentTime.
• Return resultIndex.

We initialize longestDuration to -1, resultIndex to -1, and previousTime to 0. This avoids special handling for the first element and makes the loop uniform.

We then iterate through the events array. In each iteration, we calculate the current push duration by subtracting previousTime from the current event's time.

We compare this currentDuration with longestDuration:

• If currentDuration is strictly greater than longestDuration, it means we've found a new longest push. We update longestDuration to currentDuration and resultIndex to the current button's index.
• If currentDuration is equal to longestDuration, we have a tie. According to the problem, we should choose the button with the smaller index. So, we update resultIndex to the minimum of its current value and the current button's index.

After checking, we update previousTime to the current event's time to prepare for the next iteration. After the loop finishes, resultIndex will hold the index of the button with the longest push time, with ties correctly resolved.

1class Solution {2    public int longestPush(int[][] events) {3        int longestDuration = -1;4        int resultIndex = -1;5        int previousTime = 0;6 7        for (int[] event : events) {8            int index = event[0];9            int time = event[1];10            int duration = time - previousTime;11 12            if (duration > longestDuration) {13                longestDuration = duration;14                resultIndex = index;15            } else if (duration == longestDuration) {16                if (index < resultIndex) {17                    resultIndex = index;18                }19            }20            previousTime = time;21        }22        return resultIndex;23    }24}