Approach #1: Next Array [Accepted]

Intuition

The problem statement asks us to find the next occurrence of a warmer temperature. Because temperatures can only be in [30, 100], if the temperature right now is say, T[i] = 50, we only need to check for the next occurrence of 51, 52, ..., 100 and take the one that occurs soonest.

Algorithm

Let's process each i in reverse (decreasing order). At each T[i], to know when the next occurrence of say, temperature 100 is, we should just remember the last one we've seen, next[100].

Then, the first occurrence of a warmer value occurs at warmer_index, the minimum of next[T[i]+1], next[T[i]+2], ..., next[100].

Complexity Analysis

  • Time Complexity: , where is the length of T and is the number of allowed values for T[i]. Since , we can consider this complexity .

  • Space Complexity: , the size of the answer and the next array.


Approach #2: Stack [Accepted]

Intuition

Consider trying to find the next warmer occurrence at T[i]. What information (about T[j] for j > i) must we remember?

Say we are trying to find T[0]. If we remembered T[10] = 50, knowing T[20] = 50 wouldn't help us, as any T[i] that has its next warmer ocurrence at T[20] would have it at T[10] instead. However, T[20] = 100 would help us, since if T[0] were 80, then T[20] might be its next warmest occurrence, while T[10] couldn't.

Thus, we should remember a list of indices representing a strictly increasing list of temperatures. For example, [10, 20, 30] corresponding to temperatures [50, 80, 100]. When we get a new temperature like T[i] = 90, we will have [5, 30] as our list of indices (corresponding to temperatures [90, 100]). The most basic structure that will satisfy our requirements is a stack, where the top of the stack is the first value in the list, and so on.

Algorithm

As in Approach #1, process indices i in descending order. We'll keep a stack of indices such that T[stack[-1]] < T[stack[-2]] < ..., where stack[-1] is the top of the stack, stack[-2] is second from the top, and so on; and where stack[-1] > stack[-2] > ...; and we will maintain this invariant as we process each temperature.

After, it is easy to know the next occurrence of a warmer temperature: it's simply the top index in the stack.

Here is a worked example of the contents of the stack as we work through T = [73, 74, 75, 71, 69, 72, 76, 73] in reverse order, at the end of the loop (after we add T[i]). For clarity, stack only contains indices i, but we will write the value of T[i] beside it in brackets, such as 0 (73).

  • When i = 7, stack = [7 (73)]. ans[i] = 0.
  • When i = 6, stack = [6 (76)]. ans[i] = 0.
  • When i = 5, stack = [5 (72), 6 (76)]. ans[i] = 1.
  • When i = 4, stack = [4 (69), 5 (72), 6 (76)]. ans[i] = 1.
  • When i = 3, stack = [3 (71), 5 (72), 6 (76)]. ans[i] = 2.
  • When i = 2, stack = [2 (75), 6 (76)]. ans[i] = 4.
  • When i = 1, stack = [1 (74), 2 (75), 6 (76)]. ans[i] = 1.
  • When i = 0, stack = [0 (73), 1 (74), 2 (75), 6 (76)]. ans[i] = 1.

Complexity Analysis

  • Time Complexity: , where is the length of T and is the number of allowed values for T[i]. Each index gets pushed and popped at most once from the stack.

  • Space Complexity: . The size of the stack is bounded as it represents strictly increasing temperatures.


Analysis written by: @awice.