EACH DAY WAITS FOR THE NEXT WARMER ONE. KEEP A STACK OF DAYS STILL WAITING, COLDEST ON TOP. WHEN A WARMER DAY ARRIVES IT ANSWERS EVERY COLDER DAY ON THE STACK AT ONCE, AND EACH GAP IS JUST THE INDEX DISTANCE. A DAY THAT NEVER SEES A WARMER ONE STAYS 0. EACH DAY IS PUSHED AND POPPED ONCE, SO THE WHOLE PASS IS O(N).

BLUE IS TODAY. TAN TILES ARE DAYS STILL WAITING ON THE STACK, A DECREASING RUN OF TEMPERATURES. GREEN TILES JUST GOT THEIR ANSWER WHEN TODAY BEAT THEM. THE RES ROW FILLS IN PLACE, OUT OF ORDER, LEAVING HOLES WHERE DEEPER DAYS STILL WAIT. THE STACK SHELF LISTS EACH WAITING DAY AND THE TEMPERATURE IT POINTS TO.

1def daily_temperatures(temperatures):2    stack = []3    res = [0] * len(temperatures)4    for i in range(len(temperatures)):5        t = temperatures[i]6        while stack and temperatures[stack[-1]] < t:7            j = stack.pop()8            distance = i - j9            res[j] = distance10        stack.append(i)11    return res