Minimum substring window

1. /********************************************** Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n). For example, S = "ADOBECODEBANC" T = "ABC" Minimum window is "BANC". Note: If there is no such window in S that covers all characters in T, return the empty string "". If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S. ***********************************/ class Solution { public: string minWindow(string s, string t) { int n = s.size(); int m = t.size(); //two pointers, and two hashtable unordered_map<char,int> mytable; unordered_map<char,int> currtable; int count = 0; string result = ""; int minLen = INT_MAX; for(int i = 0;i mytable[t[i]]++; int left = 0; int right = 0; for(int right = 0;right

{ char c = s[right]; //move the right pointer if(mytable.find(c) != mytable.end())//found { if(currtable[c]// make all number from t being counted count++; currtable[c]++; } if(count == m) //found { char b = s[left]; // pay more attention to the following logic, how to move left pointer. while (mytable.find(b)==mytable.end() || currtable[b]> mytable[b]) { //found the character currtable[b]--; //go to next character left++; b = s[left]; } if((right-left+1) { minLen = right-left+1; result = s.substr(left,right-left+1); } } } return result; } };

binary search solution for minimum value in sorted array

Suppose a sorted array is rotated at some pivot unknown to you beforehand.
(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).
Find the minimum element.
You may assume no duplicate exists in the array.

 class Solution {
public:
    int findMin(vector& nums) {
        int low = 0;
        int n = nums.size();
        int high = n-1;
     
      while(low=nums[high])
      {
          int mid = low + (high-low)/2;
          //notes the index
          if(nums[mid]>nums[high]) low = mid+1;  //left sorted,search in right part
          else if(nums[mid]      }
      return nums[low];
    }
};



if there is duplicates in the rotated sorted array.

class Solution {
public:
    int findMin(vector& nums) {
        int low = 0;
        int n = nums.size();
        int high = n-1;
      
      while(low=nums[high])
      {
          int mid = low + (high-low)/2;
          //notes the index 
          if(nums[mid]>nums[high]) low = mid+1;  //right sorted
          else if(nums[mid]          else low = low+1;
      }
      return nums[low];
    }
};


search insert position

class Solution {
public:
    int searchInsert(vector& nums, int target) {
        int low = 0;
        int n = nums.size();
        int high = n-1;
      
          while(low          {
              int mid = low + (high-low)/2;
              //notes the index 
              if(nums[mid]              else 
                 high = mid; //right sorted,search in left part
          }
          return nums[low]    }
};