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hello-algo/en/codes/javascript/chapter_searching/binary_search_edge.js
Yudong Jin 14c314c08c Retranslate the complete English edition from Simplified Chinese (#2010)
* Retranslate the complete English edition from Simplified Chinese

* Polish three English explanations after reader-focused review
2026-10-31 03:35:51 +08:00

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1.6 KiB
JavaScript

/**
* File: binary_search_edge.js
* Created Time: 2023-08-22
* Author: Gaofer Chou (gaofer-chou@qq.com)
*/
const { binarySearchInsertion } = require('./binary_search_insertion.js');
/* Find the leftmost occurrence of target using binary search */
function binarySearchLeftEdge(nums, target) {
// This is equivalent to finding the insertion point of target
const i = binarySearchInsertion(nums, target);
// If target was not found, return -1
if (i === nums.length || nums[i] !== target) {
return -1;
}
// Target was found; return index i
return i;
}
/* Find the rightmost occurrence of target using binary search */
function binarySearchRightEdge(nums, target) {
// Reduce the problem to finding the leftmost insertion point of target + 1
const i = binarySearchInsertion(nums, target + 1);
// j points to the rightmost target, while i points to the first element greater than target
const j = i - 1;
// If target was not found, return -1
if (j === -1 || nums[j] !== target) {
return -1;
}
// Target was found; return index j
return j;
}
/* Driver code */
// Array containing duplicate elements
const nums = [1, 3, 6, 6, 6, 6, 6, 10, 12, 15];
console.log('\nArray nums = ' + nums);
// Find the left and right boundaries using binary search
for (const target of [6, 7]) {
let index = binarySearchLeftEdge(nums, target);
console.log('Leftmost occurrence of ' + target + ' has index ' + index);
index = binarySearchRightEdge(nums, target);
console.log('Rightmost occurrence of ' + target + ' has index ' + index);
}