// Infosys
// N flowers on a Recatangular pana
int ans = 100000000;
void solve(vector<int> a, int n, int k, int index, int sum,
int maxsum)
{
if (k == 1)
{
maxsum = max(maxsum, sum);
sum = 0;
for (int i = index; i < n; i++)
{
sum += a[i];
}
maxsum = max(maxsum, sum);
ans = min(ans, maxsum);
return;
}
sum = 0;
for (int i = index; i < n; i++)
{
sum += a[i];
maxsum = max(maxsum, sum);
solve(a, n, k - 1, i + 1, sum, maxsum);
}
}
int GetMaxBeauty(int N, int K, vector<int> A)
{

solve(A, N, K, 0, 0, 0);
return ans;
}


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class CountOccurrences {
public static void main(String[] args) {
String str = "aabbccd";
char a[]=str.toCharArray();
int cnt = 0;

for (int i = 0; i < a.length; i++) {
for (int j = i + 1; j < a.length; j++)
{
if ((a[i] == a[j])) {
cnt++;
}

}
}
System.out.println(cnt);
}
}

Java

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Guys ❤️

Infosys Exam on 29/7/22 , 30/7/22,31/07/22

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Guys ❤️

Infosys Exam 3 PM Solution
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Note : Some codes are already in the channel/group, you can search the code by their names.

def minOperations(arr1, arr2, i, j):

     

    # Base Case

    if arr1 == arr2:

        return 0

         

    if i >= len(arr1) or j >= len(arr2):

        return 0

     

    # If arr[i] < arr[j]

    if arr1[i] < arr2[j]:

         

        # Include the current element

        return 1 \

        + minOperations(arr1, arr2, i + 1, j + 1)

         

    # Otherwise, excluding the current element

    return max(minOperations(arr1, arr2, i, j + 1),

               minOperations(arr1, arr2, i + 1, j))

     

# Function that counts the minimum

# moves required to sort the array

def minOperationsUtil(arr):

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    brr = sorted(arr);

     

    # If both the arrays are equal

    if(arr == brr):

         

        # No moves required

        print("0")

 

    # Otherwise

    else:

         

        # Print minimum operations required

        print(minOperations(arr, brr,)

Python
Q) minimum operations

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array AR of size N

perfect_sum(arr, s, result) :
x = [0]*len(arr)
j = len(arr) - 1

while (s > 0) :
x[j] = s % 2
s = s // 2
j -= 1
sum = 0
// https://t.me/Coding_human

for i in range(len(arr)) :
if (x[i] == 1) :
sum += arr[i]
if (sum == result) :
print("{",end="");
for i in range(len(arr)) :
if (x[i] == 1) :
print(arr[i],end= ", ");
print("}, ",end="")

def print_subset(arr, K) :
x = pow(2, len(arr))
for i in range(1, x):
perfect_sum(arr, i, K)

# Driver code
arr = [ ]
n = int(input("Enter length of array : "))
s=int(input("Enter sum : "))
for i in range(n):
ele=int(input("Enter element : "))
arr.append(ele)
print_subset(arr, s)

Python

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# A Naive recursive python program to find minimum of coins

# to make a given change V

  

import sys

  

# m is size of coins array (number of different coins)

def minCoins(coins, m, V):

  

    # base case

    if (V == 0):

        return 0

  

    # Initialize result

    res = sys.maxsize

      

    # Try every coin that has smaller value than V

    for i in range(0, m):

        if (coins[i] <= V):

            sub_res = minCoins(coins, m, V-coins[i])

  

            # Check for INT_MAX to avoid overflow and see if

            # result can minimized

            if (sub_res != sys.maxsize and sub_res + 1 < res):

                res = sub_res + 1

  // https://t.me/Coding_human

    return res

  

# Driver program to test above function

coins = [9, 6, 5, 1]

m = len(coins)

V = 11

print("Minimum coins required is",minCoins(coins, m, V))

Python 3

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Guys ❤️

Infosys Exam 3 PM Solution
Telegram:- https://t.me/infosys_sp_dsp_all_code
https://t.me/Coding_human

Note : Some codes are already in the channel/group, you can search the code by their names.

C++
Lexicographicallysmallest

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#include<bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
string str;
cin >> str;
int n = str.length(), ans = 0;
for(int i = 0; i < n; i++){
if(str[i] == '5' || str[i] == '6')
continue;
else
ans += 1;
}
cout << ans << endl;
}
return 0;
}

Beautiful Number

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// Word Subset
class Solution:
def wordSubsets(self, words1: List[str], words2: List[str]) -> List[str]:
ans = set(words1)
letters = {}
for i in words2:
for j in i:
count = i.count(j)
if j not in letters or count > letters[j]:
letters[j] = count
for i in words1:
for j in letters:
if i.count(j) < letters[j]:
ans.remove(i)
break
return list(ans)

Python 3

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#include <bits/stdc++.h>
using namespace std;
#define ll long long int
ll solve(ll n)
{


ll position = 1;
ll m = 1;

while (!(n & m)) {


m = m << 1;
position++;
}
return position;
}
int main() {
// your code goes here
ll n,p;
cin>>n;
vector<int> ans(n);
for(int i=0;i<n;i++)
{
cin>>p;

ans[i] = solve(p+1);

}
for(int i=0;i<n;i++)
cout<<ans[i]<<"\n";
return 0;
}

C++

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⭕L&T Technical Interview Experience⭕

1) Overview of internship?
2) What you have learned from the
internship?
3) Safety measures adopted by the
company?
4) What practical exposure did you get
during the internship?
5) What is bio-fuel?
6) What is the impact of bio-fuel blending
on the economy? (I had said my
interest is reading business news)
7) Do you know the process of biofuel
manufacturing?
8) Explain "Fuel and
combustion" (Elective subject).
9) Why excess air is required during
combustion?
10) What is the half-burned coal is
called? What will you do with that?

Note: No questions were asked from
the final year project. Majority Of
Questions (Internship, Elective
Subject, Hobbies).

Tips: 1) Read thoroughly the internship
report.
2) Basics of MTO, HTO, FFO, CET are
important.
3) Knowledge about software like
DWSIM/ASPAN can give you an
upper edge.

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Jai Hind 🇮🇳

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Tina code
Python

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