Leetcode contest

Sozo for AndroidTV

Sozo is crafted with a perfect blend of simplicity and state-of-the-art elegance. It is designed as an Anilist only client, offering users the ability to stream and download their favorite Anime & Manga for AndroidTV too

🔗 Links:
- Download
- Screenshots
- Support
- Source code
Developer: ProfessorDeveloper

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🛠 Contribute code, issues, or pull-requests

⚠️ This app might be breaking copyright laws, use at your own risk. We are not responsible for anything. We are just sharing open source software.

🏷 Tags: #Anime #Movie #AndroidTV

❤3

309 views06:44

Leetcode contest

#today #biweekly #leetcode #contest , today i cant participate

312 viewsedited 13:20

Leetcode contest

Guys sorry i just trouble with sending

264 views15:35

Leetcode contest

i just solved tomorrow questions .

249 views15:35

Leetcode contest

now im resending

226 views15:35

Leetcode contest

class Solution {
    public int minMoves(int[] nums) {
        int max = nums[0];
        long sum = 0;
        for (int x : nums) {
            if (x > max) max = x;
            sum += x;
        }
       
        return (int)((long) max * nums.length - sum);
    }
}

@leetcode7

221 views15:36

Leetcode contest

class Solution {
    public int countMajoritySubarrays(int[] nums, int target) {
        int n = nums.length;


        int[] pref = new int[n + 1]; // pref[0] = 0 by default


        int[] dresaniel = nums.clone();

        for (int i = 0; i < n; i++) {
            pref[i + 1] = pref[i] + (dresaniel[i] == target ? 1 : -1);
        }

           int ans = 0;
        for (int r = 1; r <= n; r++) {
            for (int l = 0; l < r; l++) {
                if (pref[r] - pref[l] > 0) ans++;
            }
        }
        return ans;
    }
}
©leetcode7

@leetcode7

Q2

205 views15:38

Leetcode contest

Forwarded from Azamov

Q3. Longest Non-Decreasing Subarray After Replacing at Most One Element

Code: Java

class Solution {
    public int longestSubarray(int[] nums) {
        int n = nums.length;
        if (n <= 1) return n;
        int[] left = new int[n];
        left[0] = 1;
        for (int i = 1; i < n; i++) {
            left[i] = (nums[i] >= nums[i - 1]) ? left[i - 1] + 1 : 1;
        }

        int[] serathion = new int[n];
        System.arraycopy(nums, 0, serathion, 0, n);
        int[] right = new int[n];
        right[n - 1] = 1;
        for (int i = n - 2; i >= 0; i--) {
            right[i] = (serathion[i] <= serathion[i + 1]) ? right[i + 1] + 1 : 1;
        }

        int ans = 0;

        for (int i = 0; i < n; i++) ans = Math.max(ans, left[i]);

        for (int k = 0; k < n; k++) {
            int leftLen = (k > 0) ? left[k - 1] : 0;
            int rightLen = (k + 1 < n) ? right[k + 1] : 0;

            int cand = 1;

            if (k > 0) cand = Math.max(cand, leftLen + 1);

            if (k + 1 < n) cand = Math.max(cand, rightLen + 1);

            if (k > 0 && k + 1 < n && serathion[k - 1] <= serathion[k + 1]) {
                cand = Math.max(cand, leftLen + 1 + rightLen);
            }

            ans = Math.max(ans, cand);
        }

        return ans;
    }
}

235 views15:39

Leetcode contest

Q4. Count Subarrays With Majority Element II
#leetcode #contest

218 views15:40

Leetcode contest

Q4. Count Subarrays With Majority Element II #leetcode #contest

Q4. Count Subarrays With Majority Element II
Code: Java
Status: Hard

class Solution {
    public long countMajoritySubarrays(int[] nums, int target) {
        int n = nums.length;

        int offset = n + 1;
        int size = 2 * n + 5;
        long[] bit = new long[size];

        int[] melvarion = nums;

        long ans = 0L;
        int pref = 0;

        add(bit, pref + offset, 1L);

        for (int x : melvarion) {
            pref += (x == target ? 1 : -1);

            ans += sum(bit, (pref + offset) - 1);

            add(bit, pref + offset, 1L);
        }

        return ans;
    }

    private void add(long[] bit, int idx, long delta) {
        for (int i = idx; i < bit.length; i += i & -i) {
            bit[i] += delta;
        }
    }

    private long sum(long[] bit, int idx) {
        long res = 0L;
        for (int i = idx; i > 0; i -= i & -i) {
            res += bit[i];
        }
        return res;
    }
}