1st code
Python

Deloitte exam

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def check_digit_position(s):
    if s[0].isdigit() and not s[-1].isdigit():
        return "Yes"
    elif not s[0].isdigit() and s[-1].isdigit():
        return "No"
    else:
        return "Invalid input"

input_str = input().strip()

print(check_digit_position(input_str))

Python
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def check_first_and_last_char(Str):
    if Str[0].isdigit() and not Str[-1].isdigit():
        print("Yes")
    elif Str[-1].isdigit() and not Str[0].isdigit():
        print("No")
    else:
        print("Invalid input")

Str = input()
check_first_and_last_char(Str)

Python

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int solve(vector<int>& A) { int s = 0; for (const auto&n: A) { string sn = to_string(n); char md = *max_element(sn.begin().

sn.end());

s += md-'0';

}

return s;

}

C++

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L&t

Q) Encryption by digits

int solve(vector<int>& arr) { int narr.size(); if (n == 0) return 0;

int maxLen = 1;

int currLen = 1;

bool f = true;

for (int i = 1; i < n; ++i) { if ((arr[i] > arr[i-1] && f) || (arr[i] < arr[i-1] && !f)) { ++currLen; } else { maxLen = max(maxLen, currLen); currLen = 2;

f = arr[i] > arr[i-1]
}
}
maxLen = max(maxLen, currLen); return maxLen;
}



C++

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L&t

Q) Ever - increasing & ever decreasing array segments

public class Solution {

    public static int longestArraySegment(int array_length, List<Integer> arr) {
        int maxLength = 1;
        int currentLength = 1;

        for (int i = 1; i < array_length; i++) {
            if (arr.get(i) > arr.get(i - 1)) {
                currentLength++;
            } else {
                currentLength = 1;
            }
            maxLength = Math.max(maxLength, currentLength);
        }

        currentLength = 1;
        for (int i = 1; i < array_length; i++) {
            if (arr.get(i) < arr.get(i - 1)) {
                currentLength++;
            } else {
                currentLength = 1;
            }
            maxLength = Math.max(maxLength, currentLength);
        }

        return maxLength;
    }

Ever increasing and ever decreasing - L&T

Java

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int

getMinlength(vector<int>a,int k)

{

int ans=1;

long long mul=1;

for(int i=0;i<a.size();i++){

if(mul*a[i]>k){

ans++;

mul=a[i];

}

else{

mul*=a[i];

}

}

return ans;

}

L&T

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Q) compressing array

def helper(reward):

nums = []

for i in reward:

heappush(nums, -i)

ans = -heappop(nums)

i = 1

while nums:

temp = -heappop(nums)

if temp - i <= 0: break

ans += temp - i

i += 1

return ans

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