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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Infosys Exam 3 PM Solution
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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
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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

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    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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Infosys Exam 3 PM Solution
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