|
| 1 | +#### Summary |
| 2 | +A direct application of the Binary Search algorithm. |
| 3 | + |
| 4 | +#### Complexity |
| 5 | +* Time complexity: O(log n) |
| 6 | +* Space complexity: O(1) |
| 7 | + |
| 8 | +#### Template of Binary Search (in C++) |
| 9 | +``` |
| 10 | +// Return the index if the target value could be found in the array, |
| 11 | +// otherwise, return -1; |
| 12 | +int binary_search(vector<int> arr, int target) { |
| 13 | + int left = 0, right = arr.size() - 1; |
| 14 | + while(left <= right) { |
| 15 | + int mid = (right - left) / 2 + left; |
| 16 | + if (arr[mid] == target) { |
| 17 | + return mid; |
| 18 | + } |
| 19 | + else if (arr[mid] > target){ |
| 20 | + right = mid - 1; |
| 21 | + } |
| 22 | + else if (arr[mid] < target){ |
| 23 | + left = mid + 1; |
| 24 | + } |
| 25 | + } |
| 26 | + return -1; |
| 27 | +} |
| 28 | +``` |
| 29 | + |
| 30 | +#### Other variants |
| 31 | + |
| 32 | +1. Left bound. Find the first index i such that arr[i] == target, otherwise return -1. |
| 33 | + |
| 34 | +* Using half-closed interval [L, R). |
| 35 | +``` |
| 36 | +int left_bound_binary_Search(vector<int> arr, int target) { |
| 37 | + int left = 0, right = arr.size(); |
| 38 | + while(left < right) { |
| 39 | + int mid = (right - left) / 2 + left; |
| 40 | + if (arr[mid] == target) { |
| 41 | + right = mid; |
| 42 | + } |
| 43 | + else if (arr[mid] > target){ |
| 44 | + right = mid; |
| 45 | + } |
| 46 | + else if (arr[mid] < target){ |
| 47 | + left = mid + 1; |
| 48 | + } |
| 49 | + } |
| 50 | +
|
| 51 | + if (left >= arr.size()) return -1; |
| 52 | + return (arr[left] == target ? left : -1); |
| 53 | +} |
| 54 | +``` |
| 55 | + |
| 56 | +* Using closed inteval [L, R] |
| 57 | +``` |
| 58 | +int left_bound_binary_Search(vector<int> arr, int target) { |
| 59 | + int left = 0, right = arr.size() - 1; |
| 60 | + while(left <= right) { |
| 61 | + int mid = (right - left) / 2 + left; |
| 62 | + if (arr[mid] == target) { |
| 63 | + right = mid - 1; |
| 64 | + } |
| 65 | + else if (arr[mid] > target){ |
| 66 | + right = mid - 1; |
| 67 | + } |
| 68 | + else if (arr[mid] < target){ |
| 69 | + left = mid + 1; |
| 70 | + } |
| 71 | + } |
| 72 | +
|
| 73 | + if (left >= arr.size()) return -1; |
| 74 | + return (arr[left] == target ? left : -1); |
| 75 | +} |
| 76 | +``` |
| 77 | + |
| 78 | +2. Right bound. Find the last index i such that arr[i] == target, otherwise return -1. |
| 79 | + |
| 80 | +* Using half-closed interval [L, R). |
| 81 | +``` |
| 82 | +int right_bound_binary_Search(vector<int> arr, int target) { |
| 83 | + int left = 0, right = arr.size(); |
| 84 | + while(left < right) { |
| 85 | + int mid = (right - left) / 2 + left; |
| 86 | + if (arr[mid] == target) { |
| 87 | + left = mid + 1; |
| 88 | + } |
| 89 | + else if (arr[mid] > target){ |
| 90 | + right = mid - 1; |
| 91 | + } |
| 92 | + else if (arr[mid] < target){ |
| 93 | + left = mid + 1; |
| 94 | + } |
| 95 | + } |
| 96 | + if (left == 0) return -1; |
| 97 | + else return (arr[left - 1] == target ? (left - 1) : -1); |
| 98 | +} |
| 99 | +``` |
| 100 | + |
| 101 | +* Using closed inteval [L, R] |
| 102 | +``` |
| 103 | +int right_bound_binary_Search(vector<int> arr, int target) { |
| 104 | + int left = 0, right = arr.size() - 1; |
| 105 | + while(left <= right) { |
| 106 | + int mid = (right - left) / 2 + left; |
| 107 | + if (arr[mid] == target) { |
| 108 | + left = mid + 1; |
| 109 | + } |
| 110 | + else if (arr[mid] > target){ |
| 111 | + right = mid - 1; |
| 112 | + } |
| 113 | + else if (arr[mid] < target){ |
| 114 | + left = mid + 1; |
| 115 | + } |
| 116 | + } |
| 117 | +
|
| 118 | + if (right < 0) return -1; |
| 119 | + return (arr[right] == target ? right : -1); |
| 120 | +} |
| 121 | +``` |
| 122 | + |
| 123 | + |
| 124 | +These templates are also available [here](https://github.com/jinshendan/Leetcode/tree/master/Template/Maths/Binary-Search). |
| 125 | + |
| 126 | + |
| 127 | +#### Remarks |
| 128 | +* Use |
| 129 | +``` |
| 130 | +mid = (right - left) / 2 + left |
| 131 | +``` |
| 132 | +instead of |
| 133 | +``` |
| 134 | +mid = (left + right) / 2 |
| 135 | +``` |
| 136 | +to avoid any overflow in intermediare calculation. |
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