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;

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L&T

@allcoding1

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

@allcoding1

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