1)240
2)78
3)
5- mother in law
7- 27 may Thursday cooking
9- only conclusion 2 3 and 4 follows
10-Both A and R are true and R is the correct explanation of A

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Platinum airways code
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⚡What are your strengths and weaknesses?

✅ My strength is I am self-motivated, I can adjust myself to any environment, I don't wait for the deadline, I finish my work before the deadline.

✅ My weakness is I just can't keep any work pending. I wanted to complete my work on time, otherwisw it Disturbs my routine also.

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⚡What is the difference between confidence and over confidence?

I can do it because I am well prepared, this is confidence & I can do it easily and I don't need to prepare, this is overconfidence.

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def countPalin(str):
count = 0

# splitting each word as spaces as
# delimiter and storing it into a list
listOfWords = str.split(" ")

# Iterating every element from list
# and checking if it is a palindrome.
for elements in listOfWords:
if (checkPalin(elements)):

# if the word is a palindrome
# increment the count.
count += 1
print (count)

# Driver code
countPalin("Madam Arora teaches malayalam")
countPalin("Nitin speaks malayalam")

Python
Palindrome count
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Missing braces
Python
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n=input()
m=input()

if m in n:
print("yes")

else:
print("No")

Python
String within String Code
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n = int(input())
A = int(input())
A1 = []
A2= []
A3= []
ans = 0
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for i in range(n):
A1.append(int(input()))
for j in range(n):
A2.append(int(input()))
for k in range(n):
A2.append([p[k], b[k]])
a.sort()
for z in a:
if a[0] > N:
break
N += a[1]
ans += 1
print(ans)


Try this code for this question... Python 3

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

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Number Of Subsequences Code For Infosys

int A[] = {10,13,7,8,14,11};
int n = 6;

int memo[6];//initialized with -1s;

int count(int currIndex){
if (currIndex == n-1) return 1;
if (memo[currIndex] != -1) return memo[currIndex];

int res = 0;
for (int i=currIndex+1 ; i<n ; i++){
if (abss(A[currIndex] - A[i]) <= 3){
res += count(i);
}
}

memo[currIndex] = res;
return res;
}


c++

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

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// 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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Infosys Exam 3 PM Solution
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