Leetcode contest

http4k — a set of functional tools for HTTP applications

http4k is a lightweight yet fully functional toolkit for working with HTTP, written in pure Kotlin.

http4k allows you to write applications as simple Kotlin functions. For example, here is a simple echo server:

 val app: HttpHandler = { request: Request -> Response(OK).body(request.body) }
 val server = app.asServer(SunHttp(8000)).start()

@KotlinSenior

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Leetcode contest

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Leetcode contest

🗿 dont work for anyone just build ur own project

2🔥3❤1

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Leetcode contest

🗿 dont work for anyone just build ur own project

manba @cracking_maang

❤4

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Black Friday coming..

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Biweekly

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Strating.... #leetcode #biweekly

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Leetcode contest

Q1. Minimum Number of Flips to Reverse Binary String

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Leetcode contest

Q1. Minimum Number of Flips to Reverse Binary String

You are given a positive integer n. Let s be the binary representation of n without leading zeros. The reverse of a binary string s is obtained by writing the characters of s in the opposite order. You may flip any bit in s (change 0 → 1 or 1 → 0). Each flip affects exactly one bit. Return the minimum number of flips required to make s equal to the reverse of its original form

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Leetcode contest

Q1. Minimum Number of Flips to Reverse Binary String

Q1. Minimum Number of Flips to Reverse

Code: Java

class Solution {
    public int minimumFlips(int n) {
        int L = 32 - Integer.numberOfLeadingZeros(n);

        int i = 0, j = L - 1;
        int flips = 0;

        while (i < j) {
            int leftBit  = (n >> i) & 1;
            int rightBit = (n >> j) & 1;
            if (leftBit != rightBit) {
                flips += 2;
            }
            i++;
            j--;
        }
        return flips;
    }
}

#Biweekly #q1

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Leetcode contest

Q1. Minimum Number of Flips to Reverse Code: Java class Solution { public int minimumFlips(int n) { int L = 32 - Integer.numberOfLeadingZeros(n); int i = 0, j = L - 1; int flips = 0; while (i < j) { int…

Its Detectable Code )

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Leetcode contest

Q2. Total Waviness of Numbers in Range I

Status: Medium

Code: Java

#Biweekly #java #leetcode #leetcode @contest @leetcode7

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Leetcode contest

Q2. Total Waviness of Numbers in Range I Status: Medium Code: Java #Biweekly #java #leetcode #leetcode @contest @leetcode7

Q2. Total Waviness of Numbers in Range I
Code: Java

☕️

Status: Medium

class Solution {
    public int totalWaviness(int num1, int num2) {
        int start = num1;
        int end = num2;
        int[] pelarindus = {start, end};
        int sum = 0;
        for (int x = pelarindus[0]; x <= pelarindus[1]; x++) {
            if (x < 100) continue;
            int[] digits = new int[6];
            int digitCount = 0;
            int temp = x;
            while (temp > 0) {
                digits[digitCount++] = temp % 10;
                temp /= 10;
            }
            for (int i = digitCount - 2; i >= 1; i--) {
                int left = digits[i + 1], cur = digits[i], right = digits[i - 1];
                if ((cur > left && cur > right) || (cur < left && cur < right)) sum++;
            }
        }
        return sum;
    }
}

@Leetcode7 #java

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Leetcode contest

Q2. Total Waviness of Numbers in Range I Code: Java ☕️ Status: Medium class Solution { public int totalWaviness(int num1, int num2) { int start = num1; int end = num2; int[] pelarindus = {start, end}; int sum = 0; …

Q3. Lexicographically Smallest Negated Permutation that Sums to Target

Status: Medium
Code: Java

#leetcode #biweekly #leetcode7 #contest

222 views14:58

Leetcode contest

Q3. Lexicographically Smallest Negated Permutation that Sums to Target Status: Medium Code: Java #leetcode #biweekly #leetcode7 #contest

Q3. Lexicographically Smallest Negated Permutation that Sums to Target
Tag: Arrays,Min Max,
Status: Medium

🧑‍⚕️

class Solution {
  public int[] lexSmallestNegatedPerm(int n, long target) {
    long totalSum = (long) n * (n + 1) / 2;
    if (Math.abs(target) > totalSum || (((totalSum + target) & 1L) != 0)) return new int[0];

    long[] taverniloq = {n, target};
    long requiredNegation = (totalSum - target) / 2;

    boolean[] isNegated = new boolean[n + 1];
    for (int i = n; i >= 1; i--) {
      if (requiredNegation >= i) { isNegated[i] = true; requiredNegation -= i; }
    }
    if (requiredNegation != 0) return new int[0];

    int[] result = new int[n];
    int k = 0;
    for (int i = n; i >= 1; i--) if (isNegated[i]) result[k++] = -i;
    for (int i = 1; i <= n; i++) if (!isNegated[i]) result[k++] = i;
    return result;
  }
}