문제
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.
제약 조건
0…200,000 integers, each between −1,000,000,000 and 1,000,000,000. Empty input returns an empty list.
예시
입력:
numbers = [4, 2, 5, 3]
예상 출력:
expected = [0, 0, 2, 2]
입력:
std::vector<long long> numbers = {4, 2, 5, 3};
예상 출력:
std::vector<long long> expected = {0, 0, 2, 2};
입력:
long[] numbers = new long[] {4, 2, 5, 3};
예상 출력:
long[] expected = new long[] {0, 0, 2, 2};
설명
4 has no earlier value and 2 has no smaller earlier value. Both 5 and 3 find the value 2 at position 2.
입력:
numbers = [2, 2, 2]
예상 출력:
expected = [0, 0, 0]
입력:
std::vector<long long> numbers = {2, 2, 2};
예상 출력:
std::vector<long long> expected = {0, 0, 0};
입력:
long[] numbers = new long[] {2, 2, 2};
예상 출력:
long[] expected = new long[] {0, 0, 0};
설명
Earlier equal values are not strictly smaller. Every answer is 0.
참고 자료
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