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Commit3962831

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‎src/main/java/g3201_3300/s3220_odd_and_even_transactions/script.sql

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# Write your MySQL query statement below
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# #Medium #2024_07_18_Time_272_ms_(100.00%)_Space_0B_(100.00%)
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# #Medium #Database #2024_07_23_Time_248_ms_(85.85%)_Space_0B_(100.00%)
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select transaction_date,
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sum(case when amount%2<>0 then amount else0 end)as odd_sum,
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sum(case when amount%2=0 then amount else0 end)as even_sumfrom transactions
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packageg3201_3300.s3222_find_the_winning_player_in_coin_game;
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// #Easy #Math #Simulation #Game_Theory #2024_07_23_Time_0_ms_(100.00%)_Space_41.6_MB_(67.81%)
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publicclassSolution {
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publicStringlosingPlayer(intx,inty) {
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booleanw =false;
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while (x >0 &&y >=4) {
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x--;
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y -=4;
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w = !w;
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}
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returnw ?"Alice" :"Bob";
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}
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}
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3222\. Find the Winning Player in Coin Game
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Easy
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You are given two**positive** integers`x` and`y`, denoting the number of coins with values 75 and 10_respectively_.
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Alice and Bob are playing a game. Each turn, starting with**Alice**, the player must pick up coins with a**total** value 115. If the player is unable to do so, they**lose** the game.
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Return the_name_ of the player who wins the game if both players play**optimally**.
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**Example 1:**
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**Input:** x = 2, y = 7
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**Output:** "Alice"
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**Explanation:**
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The game ends in a single turn:
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* Alice picks 1 coin with a value of 75 and 4 coins with a value of 10.
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**Example 2:**
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**Input:** x = 4, y = 11
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**Output:** "Bob"
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**Explanation:**
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The game ends in 2 turns:
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* Alice picks 1 coin with a value of 75 and 4 coins with a value of 10.
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* Bob picks 1 coin with a value of 75 and 4 coins with a value of 10.
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**Constraints:**
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*`1 <= x, y <= 100`
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packageg3201_3300.s3223_minimum_length_of_string_after_operations;
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// #Medium #String #Hash_Table #Counting #2024_07_23_Time_9_ms_(94.23%)_Space_46.5_MB_(38.50%)
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publicclassSolution {
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publicintminimumLength(Strings) {
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int[]freq =newint[26];
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for (inti =0;i <26;i++) {
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freq[i] =0;
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}
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for (inti =0;i <s.length();i++) {
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freq[s.charAt(i) -'a']++;
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}
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intc =0;
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for (inti :freq) {
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if (i !=0) {
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if (i %2 ==0) {
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c +=2;
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}else {
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c +=1;
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}
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}
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}
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returnc;
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}
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}
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3223\. Minimum Length of String After Operations
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Medium
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You are given a string`s`.
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You can perform the following process on`s`**any** number of times:
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* Choose an index`i` in the string such that there is**at least** one character to the left of index`i` that is equal to`s[i]`, and**at least** one character to the right that is also equal to`s[i]`.
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* Delete the**closest** character to the**left** of index`i` that is equal to`s[i]`.
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* Delete the**closest** character to the**right** of index`i` that is equal to`s[i]`.
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Return the**minimum** length of the final string`s` that you can achieve.
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**Example 1:**
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**Input:** s = "abaacbcbb"
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**Output:** 5
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**Explanation:**
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We do the following operations:
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* Choose index 2, then remove the characters at indices 0 and 3. The resulting string is`s = "bacbcbb"`.
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* Choose index 3, then remove the characters at indices 0 and 5. The resulting string is`s = "acbcb"`.
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**Example 2:**
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**Input:** s = "aa"
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**Output:** 2
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**Explanation:**
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We cannot perform any operations, so we return the length of the original string.
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**Constraints:**
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* <code>1 <= s.length <= 2 * 10<sup>5</sup></code>
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*`s` consists only of lowercase English letters.
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packageg3201_3300.s3224_minimum_array_changes_to_make_differences_equal;
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// #Medium #Array #Hash_Table #Prefix_Sum #2024_07_23_Time_4_ms_(100.00%)_Space_58.4_MB_(41.64%)
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publicclassSolution {
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publicintminChanges(int[]nums,intk) {
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int[]cm =newint[k +2];
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for (inti =0;i <nums.length /2;i++) {
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inta =Math.min(nums[i],nums[nums.length -1 -i]);
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intb =Math.max(nums[i],nums[nums.length -1 -i]);
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intd =b -a;
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if (d >0) {
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cm[0]++;
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cm[d]--;
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cm[d +1]++;
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intmax =Math.max(a,k -b) +d;
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cm[max +1]++;
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}else {
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cm[1]++;
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intmax =Math.max(a,k -a);
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cm[max +1]++;
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}
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}
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intsum =cm[0];
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intres =cm[0];
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for (inti =1;i <=k;i++) {
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sum +=cm[i];
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res =Math.min(res,sum);
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}
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returnres;
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}
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}
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3224\. Minimum Array Changes to Make Differences Equal
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Medium
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You are given an integer array`nums` of size`n` where`n` is**even**, and an integer`k`.
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You can perform some changes on the array, where in one change you can replace**any** element in the array with**any** integer in the range from`0` to`k`.
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You need to perform some changes (possibly none) such that the final array satisfies the following condition:
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* There exists an integer`X` such that`abs(a[i] - a[n - i - 1]) = X` for all`(0 <= i < n)`.
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Return the**minimum** number of changes required to satisfy the above condition.
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**Example 1:**
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**Input:** nums =[1,0,1,2,4,3], k = 4
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**Output:** 2
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**Explanation:**
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We can perform the following changes:
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* Replace`nums[1]` by 2. The resulting array is <code>nums =[1,<ins>**2**</ins>,1,2,4,3]</code>.
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* Replace`nums[3]` by 3. The resulting array is <code>nums =[1,2,1,<ins>**3**</ins>,4,3]</code>.
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The integer`X` will be 2.
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**Example 2:**
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**Input:** nums =[0,1,2,3,3,6,5,4], k = 6
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**Output:** 2
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**Explanation:**
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We can perform the following operations:
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* Replace`nums[3]` by 0. The resulting array is <code>nums =[0,1,2,<ins>**0**</ins>,3,6,5,4]</code>.
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* Replace`nums[4]` by 4. The resulting array is <code>nums =[0,1,2,0,**<ins>4</ins>**,6,5,4]</code>.
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The integer`X` will be 4.
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**Constraints:**
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* <code>2 <= n == nums.length <= 10<sup>5</sup></code>
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*`n` is even.
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* <code>0 <= nums[i] <= k <= 10<sup>5</sup></code>
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packageg3201_3300.s3225_maximum_score_from_grid_operations;
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// #Hard #Array #Dynamic_Programming #Matrix #Prefix_Sum
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// #2024_07_23_Time_21_ms_(100.00%)_Space_45.1_MB_(96.92%)
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publicclassSolution {
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publiclongmaximumScore(int[][]grid) {
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finalintn =grid.length;
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long[]dp1 =newlong[n];
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long[]dp2 =newlong[n +1];
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long[]dp3 =newlong[n +1];
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long[]dp12 =newlong[n];
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long[]dp22 =newlong[n +1];
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long[]dp32 =newlong[n +1];
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longres =0;
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for (inti =0;i <n; ++i) {
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longsum =0;
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longpre =0;
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for (int[]ints :grid) {
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sum +=ints[i];
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}
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for (intj =n -1;j >=0; --j) {
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longs2 =sum;
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dp12[j] =s2 +dp3[n];
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for (intk =0;k <=j; ++k) {
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s2 -=grid[k][i];
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longv =Math.max(dp1[k] +s2,dp3[j] +s2);
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v =Math.max(v,pre +s2);
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dp12[j] =Math.max(dp12[j],v);
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if (k ==j) {
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dp22[j] =dp32[j] =v;
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res =Math.max(res,v);
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}
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}
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if (i >0) {
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pre =Math.max(pre +grid[j][i],dp2[j] +grid[j][i]);
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}
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sum -=grid[j][i];
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}
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dp22[n] =dp32[n] =pre;
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res =Math.max(res,pre);
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for (intj =1;j <=n; ++j) {
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dp32[j] =Math.max(dp32[j],dp32[j -1]);
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}
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long[]tem =dp1;
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dp1 =dp12;
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dp12 =tem;
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tem =dp2;
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dp2 =dp22;
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dp22 =tem;
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tem =dp3;
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dp3 =dp32;
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dp32 =tem;
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}
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returnres;
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}
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}
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3225\. Maximum Score From Grid Operations
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Hard
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You are given a 2D matrix`grid` of size`n x n`. Initially, all cells of the grid are colored white. In one operation, you can select any cell of indices`(i, j)`, and color black all the cells of the <code>j<sup>th</sup></code> column starting from the top row down to the <code>i<sup>th</sup></code> row.
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The grid score is the sum of all`grid[i][j]` such that cell`(i, j)` is white and it has a horizontally adjacent black cell.
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Return the**maximum** score that can be achieved after some number of operations.
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**Example 1:**
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**Input:** grid =[[0,0,0,0,0],[0,0,3,0,0],[0,1,0,0,0],[5,0,0,3,0],[0,0,0,0,2]]
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**Output:** 11
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**Explanation:**
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![](https://assets.leetcode.com/uploads/2024/05/11/one.png)
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In the first operation, we color all cells in column 1 down to row 3, and in the second operation, we color all cells in column 4 down to the last row. The score of the resulting grid is`grid[3][0] + grid[1][2] + grid[3][3]` which is equal to 11.
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**Example 2:**
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**Input:** grid =[[10,9,0,0,15],[7,1,0,8,0],[5,20,0,11,0],[0,0,0,1,2],[8,12,1,10,3]]
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**Output:** 94
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**Explanation:**
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![](https://assets.leetcode.com/uploads/2024/05/11/two-1.png)
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We perform operations on 1, 2, and 3 down to rows 1, 4, and 0, respectively. The score of the resulting grid is`grid[0][0] + grid[1][0] + grid[2][1] + grid[4][1] + grid[1][3] + grid[2][3] + grid[3][3] + grid[4][3] + grid[0][4]` which is equal to 94.
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**Constraints:**
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*`1 <= n == grid.length <= 100`
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*`n == grid[i].length`
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* <code>0 <= grid[i][j] <= 10<sup>9</sup></code>
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packageg3201_3300.s3226_number_of_bit_changes_to_make_two_integers_equal;
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// #Easy #Bit_Manipulation #2024_07_23_Time_0_ms_(100.00%)_Space_40.5_MB_(97.19%)
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publicclassSolution {
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publicintminChanges(intn,intk) {
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if ((n |k) !=n) {
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return -1;
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}
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intcnt =0;
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while (n >0 ||k >0) {
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intbitN =n &1;
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intbitK =k &1;
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if (bitN ==1 &&bitK ==0) {
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cnt++;
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}
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n >>=1;
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k >>=1;
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}
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returncnt;
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}
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}
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3226\. Number of Bit Changes to Make Two Integers Equal
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Easy
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You are given two positive integers`n` and`k`.
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You can choose**any** bit in the**binary representation** of`n` that is equal to 1 and change it to 0.
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Return the_number of changes_ needed to make`n` equal to`k`. If it is impossible, return -1.
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**Example 1:**
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**Input:** n = 13, k = 4
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**Output:** 2
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**Explanation:**
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Initially, the binary representations of`n` and`k` are <code>n = (1101)<sub>2</sub></code> and <code>k = (0100)<sub>2</sub></code>.
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We can change the first and fourth bits of`n`. The resulting integer is <code>n = (<ins>**0**</ins>10<ins>**0**</ins>)<sub>2</sub> = k</code>.
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**Example 2:**
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**Input:** n = 21, k = 21
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**Output:** 0
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**Explanation:**
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`n` and`k` are already equal, so no changes are needed.
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**Example 3:**
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**Input:** n = 14, k = 13
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**Output:**\-1
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**Explanation:**
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It is not possible to make`n` equal to`k`.
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**Constraints:**
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* <code>1 <= n, k <= 10<sup>6</sup></code>
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packageg3201_3300.s3227_vowels_game_in_a_string;
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// #Medium #String #Math #Game_Theory #Brainteaser
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// #2024_07_23_Time_4_ms_(96.15%)_Space_45.3_MB_(96.39%)
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publicclassSolution {
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publicbooleandoesAliceWin(Strings) {
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for (inti =0;i <s.length();i++) {
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charcurr =s.charAt(i);
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if (curr =='a' ||curr =='e' ||curr =='i' ||curr =='o' ||curr =='u') {
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returntrue;
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}
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}
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returnfalse;
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}
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}

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