Insertion Position

Instead of giving -1 when search fails, give the index that the number could be inserted and maintain the order

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Example: 1,2,4,6,7,8,9

input 4 return 2

input 5 return 3 (Previously is -1)

Insertion Position

public int BinarySearch(int[] A, int target) 
    int begin = 0;
    int end = A.length ;
    while(begin < end) {
        int p = (end + begin) / 2;
        if(A[p] == target) return p;
        else if(A[p] < target) {
            begin = p + 1;
        }
        else {
            end = p;
        }
    }
    return begin;
}

Find first occurrence of a target in a sorted array

Find last occurrence of a target in a sorted array

first iteration: L = 0, R = 6, M = 3   => R = M = 3

second iteration: L = 0, R = 3, M = 1 => R = M = 1

L is neighbor of R, break;

Post processing

int binarySearch(vector<int>& a, int target){
    int left = 0, right = a.size() - 1,
    int mid;
    while(left < right - 1){
        int mid = left + (right - left)/2;
        if(a[mid] == target){
            left = mid;
        }
        else if(a[mid] < target){
            left = mid;
        }
        else {
            right = mid;
        }
    }
    if(left == target) return left;
    if(right == target) return right;
    return -1;
}

int binarySearch(vector<int>& a, int target){
    int left = 0, right = a.size() - 1,
    int mid;
    while(left < right - 1){
        int mid = left + (right - left)/2;
        if(a[mid] == target){
            right = mid;
        }
        else if(a[mid] < target){
            left = mid;
        }
        else {
            right = mid;
        }
    }
    if(right == target) return right;
    if(left == target) return left;
    return -1;
}

Search In Rotated Sorted Array

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Example: 6 7 8 9 1 2 3 4 5

Search In Rotated Sorted Array

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Search In Rotated Sorted Array

public int search(int[] A, int target) {
    int begin = 0;
    int end = A.length;
    while(begin < end) {
        int p = (end + begin) / 2;
        if(A[p] == target) return p;
        else if(A[p] > A[begin]){
            if(target >= A[begin] && target < A[p]) {
                end = p;
            }
            else begin = p + 1;
        }
        else {
            if(target > A[p] && target <= A[end - 1] ) {
                begin = p + 1;
            }
            else end = p;
        }
    }
    return -1;
}

Search In Rotated Sorted Array II

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Search In Rotated Sorted Array II

public boolean search(int[] nums, int target) {
    int begin = 0;
    int end = nums.length;
    while(begin < end) {
        int p = (begin + end) / 2;
        if(nums[p] == target) return true;
        if(nums[p] > nums[begin]) {
            if(target >= nums[begin] && target < nums[p]) {
                end = p;
            }
            else begin = p + 1;
        }
        else if(nums[p] < nums[begin]) {
            if(target > nums[p] && target <= nums[end - 1] ) {
                begin = p + 1;
            }
            else end = p;                
        }
        else {
            begin ++;
        }        
    }
    return false;
}

Sqrt(x)

How does it relate to Binary Search?

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Sqrt(x)

public int mySqrt(int x) {
    if(x <= 1) return x;
    Long begin = (long)0;
    Long end = (long)x / 2 + 1;
    while(begin < end) {
        Long p = (end + begin) / 2;
        if(p * p == x) return p.intValue();
        else if(p * p < x) {
            begin = p + 1;
        }
        else {
            end = p;
        }
    }
    return end.intValue() - 1;
}