Binary Search

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Search

The most common operation in Computer Science
Search in an Sorted Array
How to search? You can always go through the whole Array
Is there a way to search easier?

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What is Binary Search

Example: Dictionary

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Binary Search

The fastest way to search in an sorted array
Time Complexity: O(logN)

T(n) = T(n/2) + c

T(n) = T(n/2) + c

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Example Code

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 -1;
}

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Example Code 2

public int BinarySearch(int[] A, int target) 
    int begin = 0;
    int end = A.length-1;
    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 - 1;
        }
    }
    return -1;
}

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Which one is better?

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There are no major differences between each other. The only thing you need to remember is the condition in the while loop

Time Complexity

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T(n) = T(\frac{n}{2}) + k

T(n) = T(\frac{n}{2}) + k

T(n) = O(lgn)

T(n) = O(lgn)

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;
}

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;
}

Binary Search

Search

What is Binary Search

Binary Search

Example Code

Example Code 2

Which one is better?

Time Complexity

Insertion Position

Insertion Position

Search In Rotated Sorted Array

Search In Rotated Sorted Array

Search In Rotated Sorted Array

Search In Rotated Sorted Array II

Search In Rotated Sorted Array II

Sqrt(x)

How does it relate to Binary Search?

Sqrt(x)

Homework

Find Minimum in Rotated Sorted Array

Homework

Search a 2D matrix

Homework (Optional)

Divide Two Integers