chore: Simplify and optimize interpolate implementation
#8397
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MatiPl01
commented
Oct 15, 2025
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+196
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| if (x > inputRange[length - 1]) { | ||
| narrowedInput = { | ||
| leftEdgeInput: inputRange[length - 2], | ||
| rightEdgeInput: inputRange[length - 1], | ||
| leftEdgeOutput: outputRange[length - 2], | ||
| rightEdgeOutput: outputRange[length - 1], | ||
| }; |
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This can be checked right away, not only when length > 2. For length == 2 it will work the same as in the previous implementation.
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Summary
The previous implementation unnecessarily used the
O(n)loop to find the current interpolation value instead of using a simple binary search. It also Used unnecessary check if length of the input range is greater than 2 as this case doesn't have to be handled separately.