Majority Element

-
Brute Force | Using Map | 100% Success

int Solution::majorityElement(const vector<int> &A) {
    map<int,int>m;
    for(int i=0;i<A.size();i++){
        m[A[i]]++;
    }
    for(int i=0;i<A.size();i++){
        if(m[A[i]]>A.size()/2)
        return A[i];
    }
    return -1;
}

Comments

Post a Comment

Popular posts from this blog

141. Linked List Cycle

-
Question The given head, the linked list head, determines whether the linked list has a cycle. A loop exists in a linked list if there is any node in the list that can be reached again by continuously tracking the next pointer. Internally, pos is used to indicate the index of the node to which the next tail pointer is attached. Note that pos is not passed as a parameter. Return true if there is a cycle in the linked list. Otherwise, return false. Example 1: Input: head = [3,2,0,-4], pos = 1 Output: true Explanation: There is a loop in the linked list where the end connects to the 1st node (index 0). Example 2: Input: head = [1,2], pos = 0 Output: true Explanation: There is a loop in the linked list where the end connects to the 0th node. Example 3: Input: head = [1], pos = -1 Output: false Explanation: There is no cycle in the linked list. Solution class Solution { public: bool hasCycle(ListNode *head) { if(head==NULL||head->next==NULL) return false; ...
Read more

67. Add Binary

-
Add Binary   (C++) class Solution { public : string addBinary ( string a, string b ) { string ans= "" ; int carry= 0 ; while (a.size()||b.size()) { int x= 0 ,y= 0 ; if (a.size()){ x=a.back()- '0' ; a.pop_back(); } if (b.size()){ x+=b.back()- '0' ; b.pop_back(); } x+=carry; if (x== 1 ){ans+= '1' ;carry= 0 ;} if (x== 0 )ans+= '0' ; if (x== 2 ){ans+= '0' ;carry= 1 ;} if (x== 3 ){ans+= '1' ;carry= 1 ;} } if (carry)ans+= '1' ; reverse(ans.begin(),ans.end()); return ans; } };
Read more

88. Merge Sorted Array

-
C++ | Simplest Way | Brute Force class Solution { public: void merge(vector& nums1, int m, vector& nums2, int n) { int i = m- 1 , j = n- 1 , index = m+n- 1 ; while (i>= 0 && j>= 0 ){ if (nums1[i]>nums2[j]) nums1[index--] = nums1[i--]; else nums1[index--] = nums2[j--]; } while (i>= 0 ) nums1[index--] = nums1[i--]; while (j>= 0 ) nums1[index--] = nums2[j--]; } };
Read more