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Commit 81b1150

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Added tasks 371, 372, 373.
1 parent 4ad6095 commit 81b1150

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package g0301_0400.s0371_sum_of_two_integers;
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// #Medium #Top_Interview_Questions #Math #Bit_Manipulation
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public class Solution {
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public int getSum(int a, int b) {
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int ans = 0;
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int memo = 0;
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int exp = 0;
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int count = 0;
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while (count < 32) {
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int val1 = a & 1;
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int val2 = b & 1;
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int val = sum(val1, val2, memo);
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memo = val >> 1;
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val = val & 1;
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a = a >> 1;
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b = b >> 1;
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ans = ans | (val << exp);
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exp = plusOne(exp);
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count = plusOne(count);
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}
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return ans;
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}
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private int sum(int val1, int val2, int val3) {
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int count = 0;
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if (val1 == 1) {
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count = plusOne(count);
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}
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if (val2 == 1) {
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count = plusOne(count);
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}
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if (val3 == 1) {
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count = plusOne(count);
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}
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return count;
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}
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private int plusOne(int val) {
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return (-(~val));
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}
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}
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371\. Sum of Two Integers
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Medium
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Given two integers `a` and `b`, return _the sum of the two integers without using the operators_ `+` _and_ `-`.
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**Example 1:**
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**Input:** a = 1, b = 2
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**Output:** 3
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**Example 2:**
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**Input:** a = 2, b = 3
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**Output:** 5
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**Constraints:**
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* `-1000 <= a, b <= 1000`
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package g0301_0400.s0372_super_pow;
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// #Medium #Math #Divide_and_Conquer
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public class Solution {
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private static final int MOD = 1337;
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public int superPow(int a, int[] b) {
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int phi = phi(MOD);
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int arrMod = arrMod(b, phi);
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if (isGreaterOrEqual(b, phi)) {
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// Cycle has started
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// cycle starts at phi with length phi
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return exp(a % MOD, phi + arrMod);
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}
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return exp(a % MOD, arrMod);
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}
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private int phi(int n) {
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float result = n;
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for (int p = 2; p * p <= n; p++) {
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if (n % p > 0) {
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continue;
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}
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while (n % p == 0) {
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n /= p;
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}
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result *= 1.0 - 1.0 / p;
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}
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if (n > 1) {
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// if starting n was also prime (so it was greater than sqrt(n))
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result *= (1.0 - (1.0 / n));
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}
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return (int) result;
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}
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// Returns true if number in array is greater than integer named phi
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private boolean isGreaterOrEqual(int[] b, int phi) {
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int cur = 0;
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for (int j : b) {
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cur = cur * 10 + j;
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if (cur >= phi) {
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return true;
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}
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}
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return false;
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}
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// Returns number in array mod phi
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private int arrMod(int[] b, int phi) {
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int res = 0;
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for (int j : b) {
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res = (res * 10 + j) % phi;
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}
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return res;
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}
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// Binary exponentiation
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private int exp(int a, int b) {
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int y = 1;
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while (b > 0) {
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if (b % 2 == 1) {
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y = (y * a) % MOD;
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}
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a = (a * a) % MOD;
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b /= 2;
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}
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return y;
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}
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}
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372\. Super Pow
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Medium
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Your task is to calculate <code>a<sup>b</sup></code> mod `1337` where `a` is a positive integer and `b` is an extremely large positive integer given in the form of an array.
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**Example 1:**
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**Input:** a = 2, b = \[3\]
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**Output:** 8
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**Example 2:**
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**Input:** a = 2, b = \[1,0\]
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**Output:** 1024
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**Example 3:**
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**Input:** a = 1, b = \[4,3,3,8,5,2\]
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**Output:** 1
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**Example 4:**
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**Input:** a = 2147483647, b = \[2,0,0\]
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**Output:** 1198
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**Constraints:**
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* <code>1 <= a <= 2<sup>31</sup> - 1</code>
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* `1 <= b.length <= 2000`
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* `0 <= b[i] <= 9`
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* `b` doesn't contain leading zeros.
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package g0301_0400.s0373_find_k_pairs_with_smallest_sums;
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// #Medium #Array #Heap_Priority_Queue
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import java.util.ArrayList;
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import java.util.List;
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import java.util.PriorityQueue;
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public class Solution {
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private static class Node {
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long sum;
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List<Integer> al;
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int index;
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Node(int index, int num1, int num2) {
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this.sum = (long) num1 + (long) num2;
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this.al = new ArrayList<>();
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this.al.add(num1);
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this.al.add(num2);
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this.index = index;
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}
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}
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public List<List<Integer>> ksmallestPairs(int[] nums1, int[] nums2, int k) {
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PriorityQueue<Node> queue = new PriorityQueue<>((a, b) -> a.sum < b.sum ? -1 : 1);
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List<List<Integer>> res = new ArrayList<>();
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for (int i = 0; i < nums1.length && i < k; i++) {
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queue.add(new Node(0, nums1[i], nums2[0]));
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}
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for (int i = 1; i <= k && !queue.isEmpty(); i++) {
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Node cur = queue.poll();
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res.add(cur.al);
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int next = cur.index;
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int lastNum1 = cur.al.get(0);
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if (next + 1 < nums2.length) {
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queue.add(new Node(next + 1, lastNum1, nums2[next + 1]));
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}
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}
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return res;
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}
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}
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373\. Find K Pairs with Smallest Sums
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Medium
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You are given two integer arrays `nums1` and `nums2` sorted in **ascending order** and an integer `k`.
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Define a pair `(u, v)` which consists of one element from the first array and one element from the second array.
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Return _the_ `k` _pairs_ <code>(u<sub>1</sub>, v<sub>1</sub>), (u<sub>2</sub>, v<sub>2</sub>), ..., (u<sub>k</sub>, v<sub>k</sub>)</code> _with the smallest sums_.
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**Example 1:**
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**Input:** nums1 = \[1,7,11\], nums2 = \[2,4,6\], k = 3
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**Output:** \[\[1,2\],\[1,4\],\[1,6\]\]
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**Explanation:** The first 3 pairs are returned from the sequence: \[1,2\],\[1,4\],\[1,6\],\[7,2\],\[7,4\],\[11,2\],\[7,6\],\[11,4\],\[11,6\]
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**Example 2:**
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**Input:** nums1 = \[1,1,2\], nums2 = \[1,2,3\], k = 2
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**Output:** \[\[1,1\],\[1,1\]\]
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**Explanation:** The first 2 pairs are returned from the sequence: \[1,1\],\[1,1\],\[1,2\],\[2,1\],\[1,2\],\[2,2\],\[1,3\],\[1,3\],\[2,3\]
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**Example 3:**
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**Input:** nums1 = \[1,2\], nums2 = \[3\], k = 3
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**Output:** \[\[1,3\],\[2,3\]\]
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**Explanation:** All possible pairs are returned from the sequence: \[1,3\],\[2,3\]
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**Constraints:**
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* <code>1 <= nums1.length, nums2.length <= 10<sup>5</sup></code>
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* <code>-10<sup>9</sup> <= nums1[i], nums2[i] <= 10<sup>9</sup></code>
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* `nums1` and `nums2` both are sorted in **ascending order**.
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* `1 <= k <= 1000`
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package g0301_0400.s0371_sum_of_two_integers;
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import static org.hamcrest.CoreMatchers.equalTo;
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import static org.hamcrest.MatcherAssert.assertThat;
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import org.junit.jupiter.api.Test;
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class SolutionTest {
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@Test
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void getSum() {
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assertThat(new Solution().getSum(1, 2), equalTo(3));
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}
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@Test
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void getSum2() {
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assertThat(new Solution().getSum(2, 3), equalTo(5));
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}
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}
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package g0301_0400.s0372_super_pow;
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import static org.hamcrest.CoreMatchers.equalTo;
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import static org.hamcrest.MatcherAssert.assertThat;
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import org.junit.jupiter.api.Test;
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class SolutionTest {
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@Test
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void superPow() {
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assertThat(new Solution().superPow(2, new int[] {3}), equalTo(8));
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}
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@Test
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void superPow2() {
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assertThat(new Solution().superPow(2, new int[] {1, 0}), equalTo(1024));
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}
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@Test
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void superPow3() {
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assertThat(new Solution().superPow(1, new int[] {4, 3, 3, 8, 5, 2}), equalTo(1));
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}
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@Test
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void superPow4() {
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assertThat(new Solution().superPow(2147483647, new int[] {2, 0, 0}), equalTo(1198));
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}
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}
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package g0301_0400.s0373_find_k_pairs_with_smallest_sums;
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import static org.hamcrest.CoreMatchers.equalTo;
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import static org.hamcrest.MatcherAssert.assertThat;
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import java.util.ArrayList;
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import java.util.Arrays;
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import org.junit.jupiter.api.Test;
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class SolutionTest {
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@Test
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void ksmallestPairs() {
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assertThat(
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new Solution().ksmallestPairs(new int[] {1, 7, 11}, new int[] {2, 4, 6}, 3),
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equalTo(
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new ArrayList<>(
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Arrays.asList(
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Arrays.asList(1, 2),
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Arrays.asList(1, 4),
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Arrays.asList(1, 6)))));
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}
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@Test
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void ksmallestPairs2() {
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assertThat(
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new Solution().ksmallestPairs(new int[] {1, 1, 2}, new int[] {1, 2, 3}, 2),
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equalTo(new ArrayList<>(Arrays.asList(Arrays.asList(1, 1), Arrays.asList(1, 1)))));
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}
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@Test
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void ksmallestPairs3() {
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assertThat(
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new Solution().ksmallestPairs(new int[] {1, 2}, new int[] {3}, 3),
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equalTo(new ArrayList<>(Arrays.asList(Arrays.asList(1, 3), Arrays.asList(2, 3)))));
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}
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}

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