MindTree Coding Questions are mix of all basic programming concepts, Dynamic Programming, data structures and algorithm etc.
In exam, you will get 2 different questions based on the above concepts such as array, string, normal mathematics.
In Mindtree, coding test section is described as:-
MindTree Coding Ques are based on the following languages.
Important instructions for Mindtree Coding Test 2024:-
| Topics | No. of Questions |
|---|---|
| Number of Questions | 2 |
| Time Limit | 45 mins |
| Difficulty | High |
| Importance | High |
Here, you can practice mindtree miscellaneous coding questions based on implementation, DS/Algo and Advanced Algo section that comes in exam. These are some of the common ques asked in mindtree coding test.
Problem Statement –
Abhijeet is one of those students who tries to get his own money by part time jobs in various places to fill up the expenses for buying books. He is not placed in one place, so what he does, he tries to allocate how much the book he needs will cost, and then work to earn that much money only. He works and then buys the book respectively. Sometimes he gets more money than he needs so the money is saved for the next book. Sometimes he doesn’t. In that time, if he has stored money from previous books, he can afford it, otherwise he needs money from his parents.
Now His parents go to work and he can’t contact them amid a day. You are his friend, and you have to find how much money minimum he can borrow from his parents so that he can buy all the books.
He can Buy the book in any order.
Function Description:
Complete the function with the following parameters:
| Name | Type | Description |
| N | Integer | How many Books he has to buy that day. |
| EarnArray[ ] | Integer array | Array of his earnings for the ith book |
| CostArray[ ] | Integer array | Array of the actual cost of the ith book. |
Constraints:
Input Format:
Output Format: The minimum money he needs to cover the total expense.
Sample Input 1:
3
[3 4 2]
[5 3 4]
Sample Output 1:
3
Explanation:
At first he buys the 2nd book, which costs 3 rupees, so he saves 1 rupee. Then he buys the 1st book, that takes 2 rupees more. So he spends his stored 1 rupee and hence he needs 1 rupee more. Then he buys the last book.
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n, ans =0,sum=0;
cin>>n;
vector arr1(n),arr2(n);
for(int i=0;i<n;i++) cin>>arr2[i];
for(int i=0;i<n;i++) cin>>arr1[i];
for(int i=0;i<n;i++) arr2[i]-=arr1[i];
sort(arr2.begin(),arr2.end(),greater());
for(int i=0;i<n;i++)
{
sum+=arr2[i];
if(sum<0)
{ans+=abs(sum);sum=0;}
}
cout<<ans;
}n=int(input())
L1=[0] *n
L2=[0]*n
for i in range(n):
L2[i]=int(input())
for i in range(n):
L1[i]=int(input())
for i in range(n):
L2[i]=L2[i]-L1[i];
L2.sort()
su=0
ans=0
for i in range(n):
su=su+L2[i]
if su<0:
ans = ans + abs(su)
su=0
print(ans)Problem Statement-
Nobel Prize-winning Austrian-Irish physicist Erwin Schrödinger developed a machine and brought as many Christopher Columbus from as many parallel universes he could. Actually he was quite amused by the fact that Columbus tried to find India and got America. He planned to dig it further.
Though totally for research purposes, he made a grid of size n X m, and planted some people of America in a position (x,y) [in 1 based indexing of the grid], and then planted you with some of your friends in the (n,m) position of the grid. Now he gathered all the Columbus in 1,1 positions and started a race.
Given the values for n, m, x, y, you have to tell how many different Columbus(s) together will explore you as India for the first time.
Remember, the Columbus who will reach to the people of America, will be thinking that as India and hence wont come further.
Function Description:
Complete the markgame function in the editor below. It has the following parameter(s):
Parameters:
| Name | Type | Description |
| n | Integer | The number of rows in the grid. |
| m | Integer | The number of columns in the grid. |
| x | Integer | The American cell’s Row. |
| y | Integer | The American cell’s Column. |
Constraints:
Input Format:
Sample Cases
Sample Input 1
2
2
2
1
Sample Output 1
1
Explanation
The only way possible is (1,1) ->(2,1) -> (2,2), so the answer is 1.
#include <bits/stdc++.h>
using namespace std;
unordered_map<int,long long int> f;
long long int Fact(int n)
{
if(f[n]) return f[n];
return f[n]=n*Fact(n-1);
}
int main()
{
int n,m,x,y;
cin>>n>>m>>x>>y;
n-=1;m-=1;x-=1;y-=1;
f[0]=f[1]=1;
int p=(Fact(m+n)/(Fact(m)*Fact(n)));
int imp=((Fact(x+y)/(Fact(x)*Fact(y)))*(Fact(m-x+n-y)/(Fact(m-x)*Fact(n-y))));
cout<<p-imp;
}import math n=int(input())-1 m=int(input())-1 x=int(input())-1 y=int(input())-1 ans=math.factorial(n+m) ans=ans//(math.factorial(n)) ans=ans//(math.factorial(m)) ans1=math.factorial(x+y) ans1=ans1//(math.factorial(x)) ans1=ans1//(math.factorial(y)) x1=n-x y1=m-y ans2=math.factorial(x1+y1) ans2=ans2//(math.factorial(x1)) ans2=ans2//(math.factorial(y1)) print(ans-(ans1*ans2))
Problem Statement-
Aashay loves to go to WONDERLA , an amusement park. They are offering students who can code well with some discount. Our task is to reduce the cost of the ticket as low as possible.
They will give some k turns to remove the cost of one ticket where the cost of tickets are combined and given as string.Help Aashay in coding as he is not good in programming and get a 50% discount for your ticket.
Constraints:
Input Format for Custom Testing:
Sample Cases:
#include<iostream>
#include<bits/stdc++.h>
using namespace std;
int smallestNumber (string num, int k)
{
if (num.length () <= k)
return 0;
unordered_map < char, int >pos;
for (int i = 0; i < num.length (); i++)
{
pos[num[i]] = i;
}
string temp = num;
sort (num.begin (), num.end ());
string ans = num.substr (0, num.length () - k);
vector < int >v;
for (int i = 0; i < ans.length (); i++)
v.push_back (pos[ans[i]]);
sort (v.begin (), v.end ());
string ret;
for (int i = 0; i < v.size (); i++)
{
ret += temp[v[i]];
}
int final = stoi (ret);
return final;
}
int main ()
{
string s;
cin >> s;
int k;
cin >> k;
int ans;
cout << smallestNumber (s, k) % (int) (pow (10, 9) + 7);
return 0;
}
import sys n=input() k=int(input()) n1=len(n) if len(n)<=k: print(0) sys.exit() a='' i=0 while i<(n1-1) and k>0: if int(n[i])>int(n[i+1]): i+=1 k-=1 continue else: a+=n[i] i+=1 a+=n[i] i+=1 if k>0: a=a[:-k] if i<=(n1-1): while i<n1: a+=n[i] i+=1 print(int(a)%((10**9)+7))
Problem Statement -:
In an airport , the Airport authority decides to charge some minimum amount to the passengers who are carrying luggage with them. They set a threshold weight value, say T, if the luggage exceeds the weight threshold you should pay double the base amount. If it is less than or equal to threshold then you have to pay $1.
Function Description:
Complete the weightMachine function in the editor below. It has the following parameter(s):
Parameters:
| Name | Type | Description |
| N | Integer | number of luggage |
| T | Integer | weight of each luggage |
| weights[ ] | Integer array | threshold weight |
Returns: The function must return an INTEGER denoting the required amount to be paid.
Constraints:
Input Format for Custom Testing:
Sample Cases:
def weightMachine(N,weights,T):
amount=0
for i in weights:
amount+=1
if(i>T):
amount+=1
return amount
N=int(input())
weights=[]
for i in range(N):
weights.append(int(input()))
T=int(input())
print(weightMachine(N,weights,T))#include
long int weightMachine(long int N,long int weights[],long int T)
{
long int amount=0,i;
for(i=0;i<N;i++)
{
amount++;
if(weights[i]>T)
{
amount++;
}
}
return amount;
}
int main()
{
long int N,i,T;
scanf("%ld",&N);
long int weights[N];
for(i=0;i<N;i++)
{ scanf("%ld",&weights[i]);
}
scanf("%ld",&T); printf("%ld",weightMachine(N,weights,T));
return 0;
}
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