問題
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.
制約
0…200,000 integers, each between −1,000,000,000 and 1,000,000,000. Empty input returns an empty list.
例
入力:
numbers = [4, 2, 5, 3]
期待される出力:
expected = [0, 0, 2, 2]
入力:
std::vector<long long> numbers = {4, 2, 5, 3};
期待される出力:
std::vector<long long> expected = {0, 0, 2, 2};
入力:
long[] numbers = new long[] {4, 2, 5, 3};
期待される出力:
long[] expected = new long[] {0, 0, 2, 2};
解説
4 has no earlier value and 2 has no smaller earlier value. Both 5 and 3 find the value 2 at position 2.
入力:
numbers = [2, 2, 2]
期待される出力:
expected = [0, 0, 0]
入力:
std::vector<long long> numbers = {2, 2, 2};
期待される出力:
std::vector<long long> expected = {0, 0, 0};
入力:
long[] numbers = new long[] {2, 2, 2};
期待される出力:
long[] expected = new long[] {0, 0, 0};
解説
Earlier equal values are not strictly smaller. Every answer is 0.
参考資料
この課題はCodeUtility用に調整されています。上記の説明と制約に従ってください。