Binary Search

Posted 2020-11-02 The Tech Road

When to use binary search:

* 我们知道解空间在一个区间 [i,j]

* valid解是解空间的一个点（也可以看作是只有一个点的range）或者一个range的起始点或者结束点

* range之外的点都不是valid解

想法一般都两种：

* 直接找到解 （比如classical binary search)

* 找中间一个点，验证这个点，通过验证结果能知道valid解在左边或者右边区间

 

Template:

    int binarySearchTemplate() {
        int i = [lower_bound_of_solution], j = [upper_bound_of_solution];               // valid range is [i,j]
        // while (i + 1 < j) {    // Overflow if i is INT_MAX
        while (i < j - 1) {
            int m = i + (j - i) / 2;
            if (m is valid solution)
                return m;
            else if (solution is on right side)         // valid range will be [i,m-1] or [i,m]
                j = m - 1, or m;
            else                                        // valid range will be [m,j] or [m+1,j]
                i = m + 1, or m;
        }
        // now check if i or j is valid solution
    }

 

457. Classical Binary Search https://www.lintcode.com/en/problem/classical-binary-search/

class Solution {
public:
    /*
     * @param nums: An integer array sorted in ascending order
     * @param target: An integer
     * @return: An integer
     */
    int findPosition(vector<int> &nums, int target) {
        if (nums.size() == 0)
            return -1;
            
        int i = 0, j = nums.size()-1;   // [i,j] is the all valid range
        while (i+1 < j) {               // Loop will break if there‘re 1 or 2 items left
            int m = i + (j - i) / 2;
            if (nums[m] == target)
                return m;
            else if (target > nums[m])
                i = m + 1;
            else
                j = m - 1;
        }
        if (nums[i] == target)          // We need to check if i or j (i and j might be same) is valid
            return i;
        else if (nums[j] == target)
            return j;
        else
            return -1;
    }
};

 

35. Search Insert Position https://leetcode.com/problems/search-insert-position/description/

class Solution {
public:
    int searchInsert(vector<int>& nums, int target) {
        int i = 0, j = nums.size();     // valid range is [0,nums.size()]
        while (i+1 < j) {
            int m = i + (j - i) / 2;
            if (nums[m] == target)
                return m;
            else if (target > nums[m])
                i = m + 1;              // if target > nums[m], valid range will be [m+1,j]
            else
                j = m;                  // if target < nums[m], vlaid range will be [i,m]
        }
        if (target <= nums[i])          // only if target <= nums[i], result will be i; otherwise j.
            return i;
        return j;
    }
};

 

 

69. Sqrt(x) https://leetcode.com/problems/sqrtx/description/

class Solution {
public:
    int mySqrt(int x) {
        int i = 0, j = x;               // valid range is [0,x]
        while (i + 1 < j) {
            long m = i + (j - i) / 2;
            if (m * m == x)
                return m;
            else if (m * m > x)         // if target < m*m, valid range will be [i,m-1]
                j = m - 1;
            else                        // if target > m*m, valid range will be [m,j]
                i = m;
        }
        if (j * j <= x)                 // j is valid only when j*j <= x
            return j;
        return i;
    }
};

34. Search for a Range https://leetcode.com/problems/search-for-a-range/description/

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        if (nums.size() == 0)   return vector<int>( { -1, -1} );
        
        int low = -1;
        int i = 0, j = nums.size()-1;
        while (i + 1 < j) {
            int m = i + (j - i) / 2;
            if (nums[m] == target)
                j = m;
            else if (nums[m] > target)
                j = m - 1;
            else
                i = m + 1;
        }
        if (nums[i] == target)
            low = i;
        else if (nums[j] == target)
            low = j;
        
        if (low == -1)  return vector<int>( { -1, -1} );
        
        i = low, j = nums.size()-1;
        while (i + 1 < j) {
            int m = i + (j - i) / 2;
            if (nums[m] == target)
                i = m;
            else if (nums[m] > target)
                j = m - 1;
            // else
            //    i = m + 1;      // should never happen
        }
        if (nums[j] == target)
            return vector<int>( { low, j} );
        else
            return vector<int>( { low, i} );
    }
};

33. Search in Rotated Sorted Array https://leetcode.com/problems/search-in-rotated-sorted-array/description/

class Solution {
public:
    int search(vector<int>& nums, int target) {
        if (nums.size() == 0)   return -1;
        
        int i = 0, j = nums.size()-1;
        while (i + 1 < j) {
            int m = i + (j - i) / 2;
            if (nums[m] == target)
                return m;
            else if (nums[i] < nums[m]) 
            {
                if (nums[i] <= target && target < nums[m])
                    j = m - 1;
                else
                    i = m + 1;
            }
            else {
                if (nums[m] < target && target <= nums[j])
                    i = m + 1;
                else
                    j = m - 1;
            }
        }
        
        if (nums[i] == target)
            return i;
        else if (nums[j] == target)
            return j;
        return -1;
    }
};

81. Search in Rotated Sorted Array II https://leetcode.com/problems/search-in-rotated-sorted-array-ii/description/

class Solution {
public:
    bool search(vector<int>& nums, int target) {
        if (nums.size() == 0)   return false;
        
        int i = 0, j = nums.size()-1;
        while (i + 1 < j) {
            int m = i + (j - i) / 2;
            if (nums[m] == target)
                return true;
            else if (nums[i] < nums[m]) 
            {
                if (nums[i] <= target && target < nums[m])
                    j = m - 1;
                else
                    i = m + 1;
            }
            else if (nums[i] > nums[m]) {
                if (nums[m] < target && target <= nums[j])
                    i = m + 1;
                else
                    j = m - 1;
            }
            else {  // nums[i] == nums[m]
                i++;
            }
        }
        
        return (nums[i] == target || nums[j] == target);
    }
};

153. Find Minimum in Rotated Sorted Array https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/description/

class Solution {
public:
    int findMin(vector<int>& nums) {     
        if (nums.size() == 0)   return -1;
        
        int i = 0, j = nums.size()-1;
        while (i + 1 < j) {
            int m = i + (j - i) / 2;
            if (nums[i] < nums[j])  // not rotated
                return nums[i];
            else if (nums[i] < nums[m]) 
                i = m + 1;
            else
                j = m;
        }
        
        return min(nums[i], nums[j]);
    }
};

154. Find Minimum in Rotated Sorted Array II https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/description/

class Solution {
public:
    int findMin(vector<int>& nums) {
        if (nums.size() == 0)   return -1;
        
        int i = 0, j = nums.size()-1;
        while (i + 1 < j) {
            int m = i + (j - i) / 2;
            if (nums[i] < nums[j])
                return nums[i];
            if (nums[i] < nums[m])
                i = m + 1;
            else if (nums[i] > nums[m])
                j = m;
            else
                i++;
        }
        
        return min(nums[i], nums[j]);
    }
};

278. First Bad Version https://leetcode.com/problems/first-bad-version/description/

// Forward declaration of isBadVersion API.
bool isBadVersion(int version);

class Solution {
public:
    int firstBadVersion(int n) {
        int i = 1, j = n;
        while (i < j - 1) {
            int m = i + (j - i) / 2;
            if (isBadVersion(m))
                j = m;
            else
                i = m+1;
        }
        
        return isBadVersion(i) ? i : j;
    }
};

378. Kth Smallest Element in a Sorted Matrix https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/description/

class Solution {
public:
    int m, n;
    int kthSmallest(vector<vector<int>>& matrix, int k) {
        m = matrix.size();
        n = matrix[0].size();
        int i = matrix[0][0], j = matrix[m-1][n-1];
        while (i < j - 1) {
            int m = i + (j - i) / 2;
            int mc = getSmallerEqualCount(matrix, m);
            if (mc >= k)
                j = m;
            else
                i = m + 1;
        }
        return getSmallerEqualCount(matrix, i) >= k ? i : j;
    }
    int getSmallerEqualCount(vector<vector<int>>& matrix, int v) {
        int res = 0;
        for (int i = 0; i < m; i++) {
            auto it = upper_bound(matrix[i].begin(), matrix[i].end(), v);
            res += it - matrix[i].begin();
        }
        return res;
    }
};