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Commit8887b82

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1 parenta2a6d20 commit8887b82

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namespaceLeetCodeNet.G0201_0300.S0289_game_of_life{
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usingXunit;
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publicclassSolutionTest{
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[Fact]
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publicvoidGameOfLife1(){
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int[][]board=newint[][]{
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newint[]{0,1,0},
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newint[]{0,0,1},
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newint[]{1,1,1},
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newint[]{0,0,0}
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};
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newSolution().GameOfLife(board);
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int[][]expected=newint[][]{
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newint[]{0,0,0},
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newint[]{1,0,1},
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newint[]{0,1,1},
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newint[]{0,1,0}
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};
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Assert.Equal(expected,board);
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}
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[Fact]
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publicvoidGameOfLife2(){
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int[][]board=newint[][]{
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newint[]{1,1},
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newint[]{1,0}
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};
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newSolution().GameOfLife(board);
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int[][]expected=newint[][]{
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newint[]{1,1},
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newint[]{1,1}
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};
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Assert.Equal(expected,board);
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}
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}
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}
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namespaceLeetCodeNet.G0201_0300.S0290_word_pattern{
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usingXunit;
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publicclassSolutionTest{
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[Fact]
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publicvoidWordPattern1(){
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Assert.True(newSolution().WordPattern("abba","dog cat cat dog"));
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}
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[Fact]
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publicvoidWordPattern2(){
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Assert.False(newSolution().WordPattern("abba","dog cat cat fish"));
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}
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[Fact]
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publicvoidWordPattern3(){
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Assert.False(newSolution().WordPattern("aaaa","dog cat cat dog"));
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}
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[Fact]
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publicvoidWordPattern4(){
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Assert.False(newSolution().WordPattern("abba","dog dog dog dog"));
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}
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}
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}
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namespaceLeetCodeNet.G0301_0400.S0373_find_k_pairs_with_smallest_sums{
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usingSystem;
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usingXunit;
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usingSystem.Collections.Generic;
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publicclassSolutionTest{
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[Fact]
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publicvoidKSmallestPairs1(){
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varresult=newSolution().KSmallestPairs(newint[]{1,7,11},newint[]{2,4,6},3);
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varexpected=newList<IList<int>>{
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newList<int>{1,2},
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newList<int>{1,4},
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newList<int>{1,6}
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};
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Assert.Equal(expected,result);
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}
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[Fact]
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publicvoidKSmallestPairs2(){
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varresult=newSolution().KSmallestPairs(newint[]{1,1,2},newint[]{1,2,3},2);
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varexpected=newList<IList<int>>{
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newList<int>{1,1},
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newList<int>{1,1}
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};
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Assert.Equal(expected,result);
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}
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[Fact]
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publicvoidKSmallestPairs3(){
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varresult=newSolution().KSmallestPairs(newint[]{1,2},newint[]{3},3);
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varexpected=newList<IList<int>>{
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newList<int>{1,3},
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newList<int>{2,3}
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};
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Assert.Equal(expected,result);
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}
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}
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}
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namespaceLeetCodeNet.G0301_0400.S0380_insert_delete_getrandom_o1{
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usingSystem;
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usingSystem.Collections.Generic;
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usingXunit;
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publicclassRandomizedSetTest{
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[Fact]
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publicvoidRandomizedSetBasic(){
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varresult=newList<string>();
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RandomizedSetrandomizedSet=null;
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result.Add("null");
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randomizedSet=newRandomizedSet();
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result.Add(randomizedSet.Insert(1).ToString().ToLower());
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result.Add(randomizedSet.Remove(2).ToString().ToLower());
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result.Add(randomizedSet.Insert(2).ToString().ToLower());
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intrandom=randomizedSet.GetRandom();
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result.Add(random.ToString());
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result.Add(randomizedSet.Remove(1).ToString().ToLower());
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result.Add(randomizedSet.Insert(2).ToString().ToLower());
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result.Add(randomizedSet.GetRandom().ToString());
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varexpected1=newList<string>{"null","true","false","true","1","true","false","2"};
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varexpected2=newList<string>{"null","true","false","true","2","true","false","2"};
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if(random==1){
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Assert.Equal(expected1,result);
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}else{
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Assert.Equal(expected2,result);
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}
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}
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}
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}
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namespaceLeetCodeNet.G0301_0400.S0383_ransom_note{
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usingSystem;
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usingXunit;
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publicclassSolutionTest{
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[Fact]
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publicvoidCanConstruct1(){
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Assert.False(newSolution().CanConstruct("a","b"));
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}
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[Fact]
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publicvoidCanConstruct2(){
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Assert.False(newSolution().CanConstruct("aa","ab"));
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}
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[Fact]
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publicvoidCanConstruct3(){
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Assert.True(newSolution().CanConstruct("aa","aab"));
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}
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}
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}
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namespaceLeetCodeNet.G0201_0300.S0289_game_of_life{
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// #Medium #Array #Matrix #Simulation #Top_Interview_150_Matrix
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// #2025_07_16_Time_0_ms_(100.00%)_Space_46.31_MB_(66.67%)
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publicclassSolution{
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publicvoidGameOfLife(int[][]board){
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intm=board.Length,n=board[0].Length;
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int[]dx={0,0,1,1,1,-1,-1,-1};
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int[]dy={1,-1,0,1,-1,0,1,-1};
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for(inti=0;i<m;i++){
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for(intj=0;j<n;j++){
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intlive=0;
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for(intd=0;d<8;d++){
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intni=i+dx[d],nj=j+dy[d];
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if(ni>=0&&ni<m&&nj>=0&&nj<n&&(board[ni][nj]&1)==1)live++;
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}
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if((board[i][j]&1)==1){
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if(live==2||live==3)board[i][j]|=2;
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}else{
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if(live==3)board[i][j]|=2;
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}
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}
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}
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for(inti=0;i<m;i++){
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for(intj=0;j<n;j++){
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board[i][j]>>=1;
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}
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}
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}
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}
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}
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289\. Game of Life
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Medium
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According to[Wikipedia's article](https://en.wikipedia.org/wiki/Conway%27s_Game_of_Life): "The**Game of Life**, also known simply as**Life**, is a cellular automaton devised by the British mathematician John Horton Conway in 1970."
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The board is made up of an`m x n` grid of cells, where each cell has an initial state:**live** (represented by a`1`) or**dead** (represented by a`0`). Each cell interacts with its[eight neighbors](https://en.wikipedia.org/wiki/Moore_neighborhood) (horizontal, vertical, diagonal) using the following four rules (taken from the above Wikipedia article):
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1. Any live cell with fewer than two live neighbors dies as if caused by under-population.
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2. Any live cell with two or three live neighbors lives on to the next generation.
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3. Any live cell with more than three live neighbors dies, as if by over-population.
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4. Any dead cell with exactly three live neighbors becomes a live cell, as if by reproduction.
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The next state is created by applying the above rules simultaneously to every cell in the current state, where births and deaths occur simultaneously. Given the current state of the`m x n` grid`board`, return_the next state_.
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**Example 1:**
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![](https://assets.leetcode.com/uploads/2020/12/26/grid1.jpg)
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**Input:** board =[[0,1,0],[0,0,1],[1,1,1],[0,0,0]]
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**Output:**[[0,0,0],[1,0,1],[0,1,1],[0,1,0]]
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**Example 2:**
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![](https://assets.leetcode.com/uploads/2020/12/26/grid2.jpg)
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**Input:** board =[[1,1],[1,0]]
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**Output:**[[1,1],[1,1]]
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**Constraints:**
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*`m == board.length`
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*`n == board[i].length`
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*`1 <= m, n <= 25`
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*`board[i][j]` is`0` or`1`.
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**Follow up:**
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* Could you solve it in-place? Remember that the board needs to be updated simultaneously: You cannot update some cells first and then use their updated values to update other cells.
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* In this question, we represent the board using a 2D array. In principle, the board is infinite, which would cause problems when the active area encroaches upon the border of the array (i.e., live cells reach the border). How would you address these problems?
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namespaceLeetCodeNet.G0201_0300.S0290_word_pattern{
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// #Easy #String #Hash_Table #Data_Structure_II_Day_7_String #Top_Interview_150_Hashmap
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// #2025_07_16_Time_0_ms_(100.00%)_Space_40.84_MB_(71.08%)
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publicclassSolution{
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publicboolWordPattern(stringpattern,strings){
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Dictionary<char,string>dict=new();
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string[]str=s.Split(" ");
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if(pattern.Length!=str.Length){
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returnfalse;
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}
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for(inti=0;i<pattern.Length;i++){
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if(!dict.ContainsKey(pattern[i])&&!dict.ContainsValue(str[i])){
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dict.Add(pattern[i],str[i]);
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}
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if(!dict.ContainsKey(pattern[i])||!str[i].Equals(dict[pattern[i]])){
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returnfalse;
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}
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}
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returntrue;
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}
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}
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}
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290\. Word Pattern
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Easy
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Given a`pattern` and a string`s`, find if`s` follows the same pattern.
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Here**follow** means a full match, such that there is a bijection between a letter in`pattern` and a**non-empty** word in`s`.
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**Example 1:**
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**Input:** pattern = "abba", s = "dog cat cat dog"
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**Output:** true
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**Example 2:**
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**Input:** pattern = "abba", s = "dog cat cat fish"
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**Output:** false
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**Example 3:**
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**Input:** pattern = "aaaa", s = "dog cat cat dog"
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**Output:** false
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**Example 4:**
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**Input:** pattern = "abba", s = "dog dog dog dog"
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**Output:** false
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**Constraints:**
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*`1 <= pattern.length <= 300`
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*`pattern` contains only lower-case English letters.
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*`1 <= s.length <= 3000`
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*`s` contains only lower-case English letters and spaces`' '`.
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*`s`**does not contain** any leading or trailing spaces.
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* All the words in`s` are separated by a**single space**.
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namespaceLeetCodeNet.G0301_0400.S0373_find_k_pairs_with_smallest_sums{
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// #Medium #Array #Heap_Priority_Queue #Top_Interview_150_Heap
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// #2025_07_16_Time_52_ms_(73.33%)_Space_92.81_MB_(30.00%)
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usingSystem.Collections.Generic;
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publicclassSolution{
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publicIList<IList<int>>KSmallestPairs(int[]nums1,int[]nums2,intk){
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varresult=newList<IList<int>>();
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if(nums1.Length==0||nums2.Length==0||k==0){
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returnresult;
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}
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varpq=newPriorityQueue<(inti,intj),int>();
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// Initialize with pairs from nums1[0] with each element in nums2 up to `k`
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for(intj=0;j<nums2.Length&&j<k;j++){
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pq.Enqueue((0,j),nums1[0]+nums2[j]);
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}
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while(k-->0&&pq.Count>0){
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var(i,j)=pq.Dequeue();
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result.Add(newList<int>{nums1[i],nums2[j]});
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// If there's another pair with the next element in nums1, add it
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if(i+1<nums1.Length){
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pq.Enqueue((i+1,j),nums1[i+1]+nums2[j]);
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}
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}
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returnresult;
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}
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}
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}

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