Problem
For each number, find the closest earlier position containing a strictly smaller number. Equal numbers do not qualify. Positions start at 1; use 0 when there is no smaller number to the left.
Implement nearest_smaller_value(numbers).
Input
numbers: Numbers in their original order.
Output
Return one earlier position (or 0) for every input value.
Einschränkungen
0…200,000 integers, each between −1,000,000,000 and 1,000,000,000. Empty input returns an empty list.
Beispiele
Eingabe:
numbers = [4, 2, 5, 3]
Erwartete Ausgabe:
expected = [0, 0, 2, 2]
Eingabe:
std::vector<long long> numbers = {4, 2, 5, 3};
Erwartete Ausgabe:
std::vector<long long> expected = {0, 0, 2, 2};
Eingabe:
long[] numbers = new long[] {4, 2, 5, 3};
Erwartete Ausgabe:
long[] expected = new long[] {0, 0, 2, 2};
Erläuterung
4 has no earlier value and 2 has no smaller earlier value. Both 5 and 3 find the value 2 at position 2.
Eingabe:
numbers = [2, 2, 2]
Erwartete Ausgabe:
expected = [0, 0, 0]
Eingabe:
std::vector<long long> numbers = {2, 2, 2};
Erwartete Ausgabe:
std::vector<long long> expected = {0, 0, 0};
Eingabe:
long[] numbers = new long[] {2, 2, 2};
Erwartete Ausgabe:
long[] expected = new long[] {0, 0, 0};
Erläuterung
Earlier equal values are not strictly smaller. Every answer is 0.
Referenzen
Diese Aufgabe wurde für CodeUtility angepasst. Beachte die oben genannten Vorgaben und Einschränkungen.