Exclusive Time of Functions LeetCode solution in Java

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class Solution {
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  // Time Complexity: O(N) for looping through logs
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  // Space Complexity: O(N) for stack
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  public int[] exclusiveTime(int n, List<String> logs) {
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    if (n == 0) {
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      return new int[0];
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    }
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    int[] result = new int[n];
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    Stack<Pair<Integer, Integer>> stack = new Stack<>();
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    for (String s : logs) {
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      String[] sArr = s.split(":");
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      int functionId = Integer.parseInt(sArr[0]);
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      String startOrEnd = sArr[1];
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      int timestamp = Integer.parseInt(sArr[2]);
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      if (startOrEnd.equals("start")) {
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        // calculate previous in-progress length
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        if (!stack.empty()) {
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          Pair<Integer, Integer> pair = stack.peek();
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          int oldFunctionId = pair.getKey();
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          int oldTimestamp = pair.getValue();
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          result[oldFunctionId] += timestamp - oldTimestamp;
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        }
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        // add new start
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        stack.push(new Pair(functionId, timestamp));
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      } else {
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        // calculate current length
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        Pair<Integer, Integer> pair = stack.pop();
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        int oldTimestamp = pair.getValue();
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        result[functionId] += timestamp - oldTimestamp + 1;
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        // reset previous function's start
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        if (!stack.empty()) {
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          pair = stack.pop();
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          Pair<Integer, Integer> replacementPair = new Pair(pair.getKey(), timestamp + 1);
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          stack.push(replacementPair);
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        }
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      }
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    }
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    return result;
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  }
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}