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Commit 4fc6f8c

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3 files changed

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+149
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‎src/revision/KthSymbolGram.java

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package revision;
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// https://leetcode.com/problems/k-th-symbol-in-grammar/
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public class KthSymbolGram {
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// here logic involves finding the parent in the previous row.
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// look this ques. in term of tree node with left child at(2*i+1 odd)
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// and right child at (2*i+2 even) .
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// now if k is odd it will be left child of prev row node parent and otherwise right child
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// if 0 then 01 and if 1 then 10 now find the parent node and then check if 1 or 0.
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public static int method(int n, int k) {
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if (n == 1 && k == 1) {
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return 0;
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}
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int prevRowParent = method(n - 1, (k + 1) / 2);
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if ((k % 2 == 0 && prevRowParent == 0) || (k % 2 == 1 && prevRowParent == 1)) {
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return 1;
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}
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return 0;
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}
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public static void main(String[] args) {
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System.out.println(method(5, 4));
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}
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}

‎src/revision/Tribonacci.java

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package revision;
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import java.util.HashMap;
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public class Tribonacci {
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public int tribonacci(int n) {
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if (n == 0) {
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return n;
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}
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if (n <= 2) {
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return 1;
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}
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int a = 0, b = 1, c = 1, d;
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for (int i = 3; i <= n; i++) {
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d = a + b + c;
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a = b;
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b = c;
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c = d;
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}
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return c;
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}
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private static HashMap<Integer, Integer> result = null;
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public int tribonacci2(int n) {
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if (result == null) {
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result = new HashMap<>();
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}
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if (n <= 0) {
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return 0;
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} else if (n == 1) {
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return 1;
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}
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if (result.containsKey(n)) {
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return result.get(n);
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}
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int valu = tribonacci2(n - 1) + tribonacci2(n - 2) + tribonacci2(n - 3);
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result.put(n, valu);
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return valu;
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}
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int trib[] = new int[38];
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public int tribonacci3(int n) {
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if (n == 0)
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return 0;
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else if (n <= 2)
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return 1;
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if (trib[n] != 0)
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return trib[n];
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int val1 = tribonacci3(n - 1);
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int val2 = tribonacci3(n - 2);
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int val3 = tribonacci3(n - 3);
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trib[n - 1] = val1;
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trib[n - 2] = val2;
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trib[n - 3] = val3;
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return val1 + val2 + val3;
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}
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}
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public static void main(String[] args) {
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}
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}

‎src/revision/incrementquery.java

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package revision;
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import java.util.Scanner;
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// given an empty array and given queries {start index,endIndex}
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// you have to increment the subarray btw the given start and end
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// brute : O(N*queries) but optimised : O(N+K)
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public class incrementquery {
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public static void main(String[] args) {
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int n = 1000;
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int arr[] = new int[n];
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Scanner sc = new Scanner(System.in);
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int q = sc.nextInt();
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int helper[] = new int[n];
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while (q-- > 0) {
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int start = sc.nextInt();
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int end = sc.nextInt();
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helper[start] = 1;
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helper[end + 1] -= 1;
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}
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arr[0] = helper[0];
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for (int i = 1; i < n; i++) {
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arr[i] = helper[i - 1] + helper[i];
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}
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for (int i = 0; i < n; i++) {
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System.out.println(arr[i] + " ");
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}
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}
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}

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