Coding problems for Expedia Inten 2021: There were 2 coding questions and 6 MCQ’s for the coding round of the Expedia 2021 Intern Round.
Question 1: A number of ways to divide objects into groups, such that no group will have fewer objects than previously formed groups?
Example:
objects=8, groups=4 Answer: 5 [1,1,1,5], [1,1,2,4], [1,1,3,3], [1,2,2,3], [2,2,2,2] Input: 8 4 Output: 5
Solution:
Simple Approach: This problem can be solved by using recursion.
// C++ program to count the // number of ways to divide objetcs in // groups. #include <bits/stdc++.h> using namespace std;
// Function to count the number // of ways to divide the number objects // in groups. int count( int pos, int prev, int objects, int groups)
{ if (pos == groups) {
if (objects == 0)
return 1;
else
return 0;
}
// if objects is divides completely
// into less than groups
if (objects == 0)
return 0;
int solution = 0;
// put all possible values
// greater equal to prev
for ( int i = prev; i <= objects; i++) {
solution += count(pos + 1, i, objects - i, groups);
}
return solution;
} // Function to count the number of // ways to divide into objects int WaysToGo( int objects, int groups)
{ return count(0, 1, objects, groups);
} // Main Code int main()
{ int objects, groups;
objects = 8;
groups = 4;
cout << WaysToGo(objects, groups);
return 0;
} |
Time complexity: O(ObjectGroups)
Dynamic Approach: The above approach will fail as time complexity will exceed, so we will apply Dynamic Programming.
// C++ implementation to count the // number of ways to divide objects in // groups. #include <bits/stdc++.h> using namespace std;
// DP 3DArray int dp[500][500][500];
// Function to count the number // of ways to divide the objects // in groups. int count( int pos, int prev, int objects, int groups)
{ // Base Case
if (pos == groups) {
if (left == 0)
return 1;
else
return 0;
}
// if objects is divides completely
// into groups
if (objects == 0)
return 0;
// If the subproblem has been
// solved, use the value
if (dp[pos][prev][objects] != -1)
return dp[pos][prev][objects];
int solution = 0;
// put all possible values
// greater equal to prev
for ( int i = prev; i <= objects; i++) {
solution += count(pos + 1, i, objects - i, groups);
}
return dp[pos][prev][objects] = solution;
} // Function to count the number of // ways to divide into groups int WaystoDivide( int objects, int groups)
{ // Initialize DP Table as -1
memset (dp, -1, sizeof (dp));
return count(0, 1, objects, groups);
} // Main Code int main()
{ int objects, groups;
objects = 8;
groups = 4;
cout << WaystoDivide(objects, groups);
return 0;
} |
Question 2: Minimum number of distinct elements after removing m items
Given an array of items, and i-th index element denotes the item id’s and given a number m, the task is to remove m elements such that there should be minimum distinct id’s left. Print the number of distinct IDs.
Examples:
Input : arr[] = { 2, 2, 1, 3, 3, 3} m = 3 Output : 1
Remove 1 and both 2's.So, only 3 will be left that's why distinct id is 1. Input : arr[] = { 2, 4, 1, 5, 3, 5, 1, 3} m = 2 Output : 3 Remove 2 and 4 completely. So, remaining ids are 1, 3 and 5 i.e. 3
#include <bits/stdc++.h> using namespace std;
// Function to find distintc id's int distIds( int a[], int n, int m)
{ unordered_map< int , int > mi;
vector<pair< int , int > > v;
int count = 0;
// Store the occurrence of ids
for ( int i = 0; i < n; i++)
mi[a[i]]++;
// Store into the vector second as first and vice-versa
for ( auto it = mi.begin(); it != mi.end(); it++)
v.push_back(make_pair(it->second, it->first));
// Sort the vector
sort(v.begin(), v.end());
int size = v.size();
// Start removing elements from the beginning
for ( int i = 0; i < size; i++) {
// Remove if current value is less than
// or equal to m
if (v[i].first <= m) {
m -= v[i].first;
count++;
}
// Return the remaining size
else
return size - count;
}
return size - count;
} // Main code int main()
{ int arr[] = { 2, 3, 1, 2, 3, 3 };
int n = sizeof (arr) / sizeof (arr[0]);
int m = 3;
cout << distIds(arr, n, m);
return 0;
} |
Time Complexity: O(n log n)