Code Blogs

 https://www.interviewbit.com/old/problems/seats/ int Solution::seats(string A) { for(int i=0;i<A.size();i++){ if(A[i]=='x') pos.push_back(i); } int n=pos.size(); long long int ans=0; int mid=n/2;int k=1; for(int i=mid-1;i>=0;i--){ ans=(ans+pos[mid]-k-pos[i])%10000003; k++; } k=1; for(int i=mid+1;i<n;i++){ ans=(ans+pos[i]-(pos[mid]+k))%10000003; k++; } return ans%10000003; } ...

 https://www.interviewbit.com/old/problems/atoi/ while (str[i] == ' ' ) { i++; } if (str[i] == ' - ' || str[i] == ' + ' ) { sign = (str[i++] == ' - ' ) ? - 1 : 1 ; } while (str[i] >= ' 0 ' && str[i] <= ' 9 ' ) { if (base > INT_MAX / 10 || (base == INT_MAX / 10 && str[i] - ' 0 ' > 7 )) { if (sign == 1 ) return INT_MAX; else return INT_MIN; } base = 10 * base + (str[i++] - ' 0 ' ); } return base * sign; int sign = 1 , base = 0 , i = 0 ; ...

 https://www.interviewbit.com/old/problems/sort-array-with-squares/      vector<int> Solution::solve(vector<int> &arr) {         int mid=0;     for(int i=0;i<arr.size();i++)     {         if(arr[i]>=0)         {             mid=i;             break;         }     }     int i=mid-1,j=mid;     vector<int>res(arr.size());     int k=0;     while(i>=0 && j<arr.size())     {         if((arr[i]*arr[i])<(arr[j]*arr[j]))         {             res...

 https://leetcode.com/problems/find-peak-element/   class Solution { public:     int findPeakElement(vector<int>& nums) {         int l=0;         int r=nums.size()-1;         while(l<r)         {             int mid=(l+r)/2;                          if(nums[mid]>nums[mid+1])                 r=mid;             else                 l=mid+1;         }         return l;  ...

  Given an integer array A of size N . You need to check that whether there exist a element which is strictly greater than all the elements on left of it and strictly smaller than all the elements on right of it. If it exists return 1 else return 0 .  https://www.interviewbit.com/old/problems/perfect-peak-of-array/ int Solution::perfectPeak(vector<int> &A) { int ans=-1; int l[A.size()]; int r[A.size()]; l[0]=A[0]; r[A.size()-1]=A[A.size()-1]; for(int i=1;i<A.size();i++){     l[i] = max(A[i],l[i-1]); } for(int i=A.size()-2;i>=0;i--){     r[i] = min(A[i],r[i+1]); } for(int i=1;i<A.size()-1;i++){     if(A[i]>l[i-1]&&A[i]<r[i+1]){         return 1;     } } if(ans==-1){     return 0; } }  

  Given a stream of numbers A . On arrival of each number, you need to increase its first occurence by 1 and include this in the stream. Return the final stream of numbers.  ip: A = [1, 1] op:  [2, 1]     vector<int> Solution::solve(vector<int> &a) {      unordered_map<int,int> m;     for(int i=0; i<a.size(); i++)     {         if(m.count(a[i]))         {             a[m[a[i]]]++;             m[a[i]+1] = m[a[i]];         }         m[a[i]] = i;     }     return a; }  

  ListNode* Solution::rotateRight(ListNode* A, int B) { ListNode* ptr = A; int len= 1 ; while (ptr-> next !=NULL){ ptr = ptr-> next ; len++; } B = B%len; if (B== 0 ) return A; ListNode* head, *temp, *start = A; int i= 1 ; ptr = A; while (i<len-B){ ptr = ptr-> next ; i++; } head = ptr-> next ; temp = head; while (temp-> next !=NULL) temp = temp-> next ; temp-> next = start; ptr-> next = NULL; return head; }

 class Solution { public:     void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {         int i = m-1, j = n-1, k = n+m-1;                  while(i>=0 && j>=0) {             if(nums1[i] >= nums2[j]) {                 nums1[k] = nums1[i];                 i--;             }             else {                 nums1[k] = nums2[j];                 j--;  ...

 string Solution::intToRoman(int A) {     int n[] = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};        string r[] = {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};     auto i = 0;     string result = "";          while (A!=0)     {         while (A>=n[i])         {             result += r[i];             A -= n[i];         }         ++i;     }     return result; }

  int Solution :: solve ( string A ) { long long v = 0 ; long long c = 0 ; long long ans = 0 ; long long mode = 1000000007 ; for ( int i = 0 ; i < A . size (); i ++ ) { if ( A [ i ] == 'a' || A [ i ] == 'e' || A [ i ] == 'i' || A [ i ] == 'o' || A [ i ] == 'u' ) { v ++ ; ans = ( ans + c ) % mode ; } else { c ++ ; ans = ( ans + v ) % mode ; } } return ( int ) ans ; }

  Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value. If target is not found in the array, return [-1, -1] .   Input: nums = [5,7,7,8,8,10], target = 8 Output: [3,4]  class Solution { public:     vector<int> searchRange(vector<int>& v, int b) {         int n=v.size();         int low=0;         int high=n-1;         int mid;         vector<int> ans(2);         ans[0]=-1;         ans[1]=-1;         while(low<=high)         {             mid=low+(high-low)/2;       ...

  You are given a license key represented as a string s that consists of only alphanumeric characters and dashes. The string is separated into n + 1 groups by n dashes. You are also given an integer k . We want to reformat the string s such that each group contains exactly k characters, except for the first group, which could be shorter than k but still must contain at least one character. Furthermore, there must be a dash inserted between two groups, and you should convert all lowercase letters to uppercase. Return the reformatted license key .   Input: s = "5F3Z-2e-9-w", k = 4 Output: "5F3Z-2E9W"   class Solution { public: string licenseKeyFormatting(string s, int k) { string ans; int n=k; for(int i=s.size()-1;i>=0;--i) { if(s[i] != '-') { if(n==0) { ans.push_back('-'); n=k; } ...

 Given a string A and integer B , remove all consecutive same characters that have length exactly B. A = "aabcd" B = 2 op: "bcd" string Solution::solve(string A, int B) { string ans=""; int i=0; int count=0; string t=""; for(i=0;i<A.length();i++){ t=t+A[i]; count++; if(A[i]!=A[i+1] ) { if(count!=B) ans=ans+t; t=""; count=0; } } return ans+t; }    

 Given an integer array A of size N , find the first repeating element in it. We need to find the element that occurs more than once and whose index of first occurrence is smallest . If there is no repeating element, return -1.   int Solution::solve(vector<int> &A) {     int n=A.size();     unordered_map<int,int> m;     if(n==1) return -1;     for(int i=0;i<n;i++)     {         m[A[i]]++;     }     for(int i=0;i<n;i++)  if(m[A[i]]>1)return A[i];     return -1; } int Solution::solve(vector<int> &A) {     int n=A.size();     int min=-1;     unordered_map<int, int> b;     for(int i=n-1;i>=0;i--){         b[A[i]] +=1;        if(b[A[i]] > 1) min=A[i]; ...

 Given the sequence of up and down steps during a hike, find and print the number of valleys walked through.  ip: 8 UDDDUDUU op: 1   _/\ _ \ / \/\/     int   countingValleys ( int   n ,   string   s )   {    int   level = 0 ;    int   valley = 0 ;       for ( auto   c :   s )    {        if ( c == 'U' )        {            if ( level ==- 1 )            {                valley ++;            }            level ++;        }        else {      ...

Given an array having both positive and negative integers. The task is to compute the length of the largest subarray with sum 0. Input: N = 8 A[] = {15,-2,2,-8,1,7,10,23} Output: 5 Explanation: The largest subarray with sum 0 will be -2 2 -8 1 7. int maxLen(int A[], int n) { int maxi=0; // store max length int sum=0; //stores the contiguous sum unordered_map<int,int>m; for(int i=0;i<n;i++) { sum+=A[i]; if(sum==0) // it is the largest sub array { maxi=i+1; } else{ if(m.find(sum)!=m.end()) //if we find the same sum that means it is a sub array with zero sum { maxi=max(maxi,i-m[sum]); } else{ m[sum]=i; //else update in hash map } } } return maxi; }      

 Given an unsorted array of integers nums , return the length of the longest consecutive elements sequence.     int longestConsecutive(vector<int>& nums) {         set<int> hashSet;               //declared a set         for(int num: nums){             hashSet.insert(num);        //inserting all array element in the set         }         int  longestStreak=0;         for(int num: nums){             if(!hashSet.count(num-1)){           //if the number does not has the previous number the update    ...

 Given an integer rowIndex , return the rowIndex th ( 0-indexed ) row of the Pascal's triangle .       vector<int> getRow(int rowIndex) {         vector<int> v(rowIndex+1,1);                for(int i=1;i<rowIndex;++i)         {             for(int j=i;j>0;--j)             {                 v[j]=v[j]+v[j-1];             }         }                  return v;     }

  Check if the product of every contiguous subsequence is different or not in a number N = 23 2 3 23 2 -> 2 3 -> 3 23 -> 6 this number is a COLORFUL number since product of every digit of a sub-sequence are different. Output : 1    int Solution::colorful(int A) {    string s= to_string(A); //convert the number to string    int n=s.size(); //size of the string    map<long long,bool> m; //map initialize    for(auto i=0;i<n;i++)  //iteration to find mul of all subsequence    {        long long int mul=1;        for(auto j=i;j<n;j++)        {            mul*=(long long int)(s[j]-'0'); //subtracted by '0' because its a string by subtracting with string it  //will be integer            if(m.find(mul)!=m.end())...

Input: numRows = 5 Output: [[1],[1,1],[1,2,1],[1,3,3,1],[1,4,6,4,1]]  //each  nth row has n numbers // first and last no is always 1                          1                     1          1               1           2           1          1         3          3            1  vector<vector<int>> generate(int numRows) {         vector<vector<int>> r(numRows);           ...

  Given a sorted array A and a target value B , return the index if the target is found. If not, return the index where it would be if it were inserted in order. You may assume no duplicates in the array.   int searchInsert ( vector < int > & A , int target ) { int n = A . size (); int start = 0 , end = n - 1 ; int mid ; while ( start <= end ){ mid = ( start + end ) / 2 ; if ( target == A [ mid ]){ return mid ; } else if ( target < A [ mid ]){ end = mid - 1 ; } else { start = mid + 1 ; } } return start ; }  

Given an sorted array A of size N . Find number of elements which are less than or equal to B .   int Solution::solve(vector<int> &A, int B) {     int start=0;     int ans=0;     int end=A.size()-1;          while(start<=end)     {         int mid=(end+start)/2;                 if(A[mid]<=B)        {            ans=mid+1;            start=mid+1;        }        else{            end=mid-1;        }     }     return ans; }

  string dayOfProgrammer( int  year) {     string res="";           if (year==1918)          {              res=  "26.09.1918\n";          }           else {               if ((year < 1918 && year%4==0)||(year%400==0) || (year%4==0&&year%100!=0))              {                  res= "12.09."+to_string(year);              }        ...

   Given an integer columnNumber , return its corresponding column title as it appears in an Excel sheet . A -> 1 B -> 2 C -> 3 ... Z -> 26 AA -> 27 AB -> 28 ...     string convertToTitle(int columnNumber ) {         string res="";         while(n--)         {             res+='A'+ (char)(n%26);             n=n/26;         }         reverse(res.begin(),res.end());         return res;     }

     vector<int> distributeCandies(int candies, int n) {         vector<int> res(n);         for (int i = 0; candies > 0; ++i) {             res[i % n] += min(candies, i + 1);             candies -= i + 1;         }         return res;     }   vector < int > distributeCandies ( int candies, int num_people) { int c = 1 , i = 0 ; vector < int > result (num_people, 0 ) ; while (candies) { int give = min(c, candies); result[i] += give; candies -= give; c++; i = (i+ 1 ) % num_people; } return result; }  

  Given an array A of positive integers,call a (contiguous,not necessarily distinct) subarray of A good if the number of different integers in that subarray is exactly B . (For example: [1, 2, 3, 1, 2] has 3 different integers 1, 2 and 3) Return the number of good subarrays of A .  input A = [1, 2, 1, 3, 4] B = 3  output 7    Subarrays formed with exactly 3 different integers: [1, 2, 1, 3], [2, 1, 3], [1, 3, 4].   int Solution::solve(vector<int> &A, int B) { unordered_map<int, int> m; int res = 0, j=0, prefix = 0, cnt=0; for(auto i = 0; i < A.size(); ++i) {     if (m[A[i]]++ == 0)         ++cnt;     if (cnt > B){         --m[A[j++]];         --cnt;         prefix = 0;     }     while (m[A[j]] > 1){     ...

 Given two binary strings a and b , return their sum as a binary string .           int i = a.size() - 1;         int j = b.size() - 1;         int carry=0;         string result = "";         while (i >= 0 || j >= 0 ) {             int sum=carry;             if (i >= 0 ) sum+=a[i--]-'0'; // its given in string to convert in integer we subtract it with '0'             if (j >= 0 ) sum+=b[j--]-'0';             carry=sum>1?1:0;             result+=to_string(sum%2);         }   ...

 Say you have an array, A , for which the i th element is the price of a given stock on day i . If you were only permitted to complete at most one transaction (i.e, buy one and sell one share of the stock), design an algorithm to find the maximum profit. Return the maximum possible profit. Input 1: A = [1, 2] Output 1: 1     int  Solution::maxProfit( const  vector< int > &arr) {      int  maxprof= 0 ;      int  minimum=INT_MAX;      for ( int  i= 0 ;i<arr.size();i++)     {         minimum=min(minimum,arr[i]);         maxprof=max(maxprof,arr[i]-minimum);              }           return  maxprof; }

Given a string s consists of some words separated by spaces, return the length of the last word in the string. If the last word does not exist, return 0 .     int lengthOfLastWord(string s) {        int ans=0;         for(int i=s.length()-1;i>=0;i--)         {             if(s[i]==' ' && ans>0) return ans;             if(s[i]!=' ' )ans++;         }         return ans;              }   int lengthOfLastWord(string s) {        int len = 0, tail = s.length() - 1;         while (tail >= 0 && s[tail] == ' ') tail--;       ...

  You are climbing a stair case. It takes n steps to reach to the top. Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top? Example : Input : 3 Return : 3 Steps : [1 1 1], [1 2], [2 1] int Solution::climbStairs(int n) { int dp[n]; if(n==0) return 1; dp[0]=1; dp[1]=2; for(int i=2;i<n;i++){ dp[i]=dp[i-1]+dp[i-2]; } return dp[n-1]; ...

 Kadane's Algorithm  int maxSubArray(vector<int>& a) {         int max_so_far = INT_MIN, max_ending_here = 0;           for (int i = 0; i < a.size(); i++)        {         max_ending_here = max_ending_here + a[i];         if (max_so_far < max_ending_here)             max_so_far = max_ending_here;           if (max_ending_here < 0)             max_ending_here = 0;        }            return max_so_far;     }