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Commit2150e4f

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packageg3301_3400.s3349_adjacent_increasing_subarrays_detection_i;
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// #Easy #Array #2024_11_15_Time_1_ms_(100.00%)_Space_44.7_MB_(18.69%)
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importjava.util.List;
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publicclassSolution {
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publicbooleanhasIncreasingSubarrays(List<Integer>nums,intk) {
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intl =nums.size();
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if (l <k *2) {
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returnfalse;
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}
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for (inti =0;i <l -2 *k +1;i++) {
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if (check(i,k,nums) &&check(i +k,k,nums)) {
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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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privatebooleancheck(intp,intk,List<Integer>nums) {
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for (inti =p;i <p +k -1;i++) {
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if (nums.get(i) >=nums.get(i +1)) {
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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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3349\. Adjacent Increasing Subarrays Detection I
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Easy
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Given an array`nums` of`n` integers and an integer`k`, determine whether there exist**two****adjacent** subarrays of length`k` such that both subarrays are**strictly****increasing**. Specifically, check if there are**two** subarrays starting at indices`a` and`b` (`a < b`), where:
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* Both subarrays`nums[a..a + k - 1]` and`nums[b..b + k - 1]` are**strictly increasing**.
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* The subarrays must be**adjacent**, meaning`b = a + k`.
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Return`true` if it is_possible_ to find**two** such subarrays, and`false` otherwise.
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**Example 1:**
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**Input:** nums =[2,5,7,8,9,2,3,4,3,1], k = 3
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**Output:** true
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**Explanation:**
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* The subarray starting at index`2` is`[7, 8, 9]`, which is strictly increasing.
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* The subarray starting at index`5` is`[2, 3, 4]`, which is also strictly increasing.
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* These two subarrays are adjacent, so the result is`true`.
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**Example 2:**
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**Input:** nums =[1,2,3,4,4,4,4,5,6,7], k = 5
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**Output:** false
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**Constraints:**
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*`2 <= nums.length <= 100`
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*`1 < 2 * k <= nums.length`
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*`-1000 <= nums[i] <= 1000`
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packageg3301_3400.s3350_adjacent_increasing_subarrays_detection_ii;
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// #Medium #Array #Binary_Search #2024_11_15_Time_10_ms_(99.76%)_Space_90.6_MB_(30.61%)
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importjava.util.List;
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publicclassSolution {
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publicintmaxIncreasingSubarrays(List<Integer>nums) {
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intn =nums.size();
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int[]a =newint[n];
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for (inti =0;i <n; ++i) {
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a[i] =nums.get(i);
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}
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intans =1;
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intpreviousLen =Integer.MAX_VALUE;
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inti =0;
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while (i <n) {
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intj =i +1;
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while (j <n &&a[j -1] <a[j]) {
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++j;
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}
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intlen =j -i;
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ans =Math.max(ans,len /2);
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if (previousLen !=Integer.MAX_VALUE) {
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ans =Math.max(ans,Math.min(previousLen,len));
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}
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previousLen =len;
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i =j;
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}
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returnans;
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}
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}
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3350\. Adjacent Increasing Subarrays Detection II
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Medium
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Given an array`nums` of`n` integers, your task is to find the**maximum** value of`k` for which there exist**two** adjacent subarrays of length`k` each, such that both subarrays are**strictly****increasing**. Specifically, check if there are**two** subarrays of length`k` starting at indices`a` and`b` (`a < b`), where:
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* Both subarrays`nums[a..a + k - 1]` and`nums[b..b + k - 1]` are**strictly increasing**.
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* The subarrays must be**adjacent**, meaning`b = a + k`.
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Return the**maximum**_possible_ value of`k`.
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A**subarray** is a contiguous**non-empty** sequence of elements within an array.
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**Example 1:**
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**Input:** nums =[2,5,7,8,9,2,3,4,3,1]
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**Output:** 3
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**Explanation:**
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* The subarray starting at index 2 is`[7, 8, 9]`, which is strictly increasing.
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* The subarray starting at index 5 is`[2, 3, 4]`, which is also strictly increasing.
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* These two subarrays are adjacent, and 3 is the**maximum** possible value of`k` for which two such adjacent strictly increasing subarrays exist.
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**Example 2:**
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**Input:** nums =[1,2,3,4,4,4,4,5,6,7]
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**Output:** 2
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**Explanation:**
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* The subarray starting at index 0 is`[1, 2]`, which is strictly increasing.
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* The subarray starting at index 2 is`[3, 4]`, which is also strictly increasing.
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* These two subarrays are adjacent, and 2 is the**maximum** possible value of`k` for which two such adjacent strictly increasing subarrays exist.
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**Constraints:**
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* <code>2 <= nums.length <= 2 * 10<sup>5</sup></code>
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* <code>-10<sup>9</sup> <= nums[i] <= 10<sup>9</sup></code>
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packageg3301_3400.s3351_sum_of_good_subsequences;
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// #Hard #Array #Hash_Table #Dynamic_Programming
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// #2024_11_15_Time_13_ms_(99.09%)_Space_55.8_MB_(68.79%)
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publicclassSolution {
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publicintsumOfGoodSubsequences(int[]nums) {
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intmax =0;
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for (intx :nums) {
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max =Math.max(x,max);
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}
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long[]count =newlong[max +3];
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long[]total =newlong[max +3];
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longmod = (int) (1e9 +7);
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longres =0;
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for (inta :nums) {
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count[a +1] = (count[a] +count[a +1] +count[a +2] +1) %mod;
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longcur =total[a] +total[a +2] +a * (count[a] +count[a +2] +1);
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total[a +1] = (total[a +1] +cur) %mod;
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res = (res +cur) %mod;
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}
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return (int)res;
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}
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}
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3351\. Sum of Good Subsequences
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Hard
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You are given an integer array`nums`. A**good** subsequence is defined as a subsequence of`nums` where the absolute difference between any**two** consecutive elements in the subsequence is**exactly** 1.
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Return the**sum** of all_possible_**good subsequences** of`nums`.
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Since the answer may be very large, return it**modulo** <code>10<sup>9</sup> + 7</code>.
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**Note** that a subsequence of size 1 is considered good by definition.
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**Example 1:**
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**Input:** nums =[1,2,1]
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**Output:** 14
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**Explanation:**
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* Good subsequences are:`[1]`,`[2]`,`[1]`,`[1,2]`,`[2,1]`,`[1,2,1]`.
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* The sum of elements in these subsequences is 14.
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**Example 2:**
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**Input:** nums =[3,4,5]
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**Output:** 40
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**Explanation:**
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* Good subsequences are:`[3]`,`[4]`,`[5]`,`[3,4]`,`[4,5]`,`[3,4,5]`.
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* The sum of elements in these subsequences is 40.
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**Constraints:**
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* <code>1 <= nums.length <= 10<sup>5</sup></code>
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* <code>0 <= nums[i] <= 10<sup>5</sup></code>
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packageg3301_3400.s3352_count_k_reducible_numbers_less_than_n;
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// #Hard #String #Dynamic_Programming #Math #Combinatorics
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// #2024_11_15_Time_11_ms_(99.58%)_Space_42.6_MB_(95.85%)
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publicclassSolution {
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privatestaticfinalintMOD = (int) (1e9 +7);
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publicintcountKReducibleNumbers(Strings,intk) {
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intn =s.length();
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int[]reducible =newint[n +1];
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for (inti =2;i <reducible.length;i++) {
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reducible[i] =1 +reducible[Integer.bitCount(i)];
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}
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long[]dp =newlong[n +1];
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intcurr =0;
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for (inti =0;i <n;i++) {
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for (intj =i -1;j >=0;j--) {
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dp[j +1] +=dp[j];
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dp[j +1] %=MOD;
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}
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if (s.charAt(i) =='1') {
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dp[curr]++;
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dp[curr] %=MOD;
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curr++;
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}
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}
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longresult =0;
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for (inti =1;i <=s.length();i++) {
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if (reducible[i] <k) {
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result +=dp[i];
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result %=MOD;
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}
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}
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return (int) (result %MOD);
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}
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}
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3352\. Count K-Reducible Numbers Less Than N
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Hard
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You are given a**binary** string`s` representing a number`n` in its binary form.
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You are also given an integer`k`.
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An integer`x` is called**k-reducible** if performing the following operation**at most**`k` times reduces it to 1:
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* Replace`x` with the**count** of set bits in its binary representation.
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For example, the binary representation of 6 is`"110"`. Applying the operation once reduces it to 2 (since`"110"` has two set bits). Applying the operation again to 2 (binary`"10"`) reduces it to 1 (since`"10"` has one set bit).
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Return an integer denoting the number of positive integers**less** than`n` that are**k-reducible**.
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Since the answer may be too large, return it**modulo** <code>10<sup>9</sup> + 7</code>.
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**Example 1:**
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**Input:** s = "111", k = 1
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**Output:** 3
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**Explanation:**
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`n = 7`. The 1-reducible integers less than 7 are 1, 2, and 4.
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**Example 2:**
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**Input:** s = "1000", k = 2
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**Output:** 6
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**Explanation:**
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`n = 8`. The 2-reducible integers less than 8 are 1, 2, 3, 4, 5, and 6.
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**Example 3:**
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**Input:** s = "1", k = 3
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**Output:** 0
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**Explanation:**
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There are no positive integers less than`n = 1`, so the answer is 0.
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**Constraints:**
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*`1 <= s.length <= 800`
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*`s` has no leading zeros.
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*`s` consists only of the characters`'0'` and`'1'`.
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*`1 <= k <= 5`
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packageg3301_3400.s3349_adjacent_increasing_subarrays_detection_i;
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importstaticorg.hamcrest.CoreMatchers.equalTo;
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importstaticorg.hamcrest.MatcherAssert.assertThat;
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importjava.util.List;
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importorg.junit.jupiter.api.Test;
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classSolutionTest {
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@Test
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voidhasIncreasingSubarrays() {
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assertThat(
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newSolution().hasIncreasingSubarrays(List.of(2,5,7,8,9,2,3,4,3,1),3),
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equalTo(true));
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}
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@Test
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voidhasIncreasingSubarrays2() {
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assertThat(
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newSolution().hasIncreasingSubarrays(List.of(1,2,3,4,4,4,4,5,6,7),5),
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equalTo(false));
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}
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}
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packageg3301_3400.s3350_adjacent_increasing_subarrays_detection_ii;
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importstaticorg.hamcrest.CoreMatchers.equalTo;
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importstaticorg.hamcrest.MatcherAssert.assertThat;
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importjava.util.List;
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importorg.junit.jupiter.api.Test;
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classSolutionTest {
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@Test
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voidmaxIncreasingSubarrays() {
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assertThat(
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newSolution().maxIncreasingSubarrays(List.of(2,5,7,8,9,2,3,4,3,1)),
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equalTo(3));
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}
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@Test
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voidmaxIncreasingSubarrays2() {
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assertThat(
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newSolution().maxIncreasingSubarrays(List.of(1,2,3,4,4,4,4,5,6,7)),
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equalTo(2));
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}
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}
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packageg3301_3400.s3351_sum_of_good_subsequences;
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importstaticorg.hamcrest.CoreMatchers.equalTo;
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importstaticorg.hamcrest.MatcherAssert.assertThat;
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importorg.junit.jupiter.api.Test;
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classSolutionTest {
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@Test
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voidsumOfGoodSubsequences() {
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assertThat(newSolution().sumOfGoodSubsequences(newint[] {1,2,1}),equalTo(14));
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}
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@Test
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voidsumOfGoodSubsequences2() {
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assertThat(newSolution().sumOfGoodSubsequences(newint[] {3,4,5}),equalTo(40));
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}
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}
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packageg3301_3400.s3352_count_k_reducible_numbers_less_than_n;
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importstaticorg.hamcrest.CoreMatchers.equalTo;
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importstaticorg.hamcrest.MatcherAssert.assertThat;
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importorg.junit.jupiter.api.Test;
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classSolutionTest {
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@Test
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voidcountKReducibleNumbers() {
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assertThat(newSolution().countKReducibleNumbers("111",1),equalTo(3));
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}
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@Test
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voidcountKReducibleNumbers2() {
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assertThat(newSolution().countKReducibleNumbers("1000",2),equalTo(6));
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}
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@Test
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voidcountKReducibleNumbers3() {
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assertThat(newSolution().countKReducibleNumbers("1",3),equalTo(0));
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}
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}

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