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feat: add Java solutions to lc problems: No.2859~2862 (doocs#1641)
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12 files changed

+273
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‎solution/2800-2899/2859.Sum of Values at Indices With K Set Bits/README.md

+11-1
Original file line numberDiff line numberDiff line change
@@ -73,7 +73,17 @@
7373
<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int sumIndicesWithKSetBits(List<Integer> nums, int k) {
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int ans = 0;
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for (int i = 0; i < nums.size(); i++) {
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if (Integer.bitCount(i) == k) {
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ans += nums.get(i);
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}
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}
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return ans;
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}
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}
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```
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### **C++**

‎solution/2800-2899/2859.Sum of Values at Indices With K Set Bits/README_EN.md

+11-1
Original file line numberDiff line numberDiff line change
@@ -65,7 +65,17 @@ Hence, the answer is nums[3] = 1.
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### **Java**
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```java
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class Solution {
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public int sumIndicesWithKSetBits(List<Integer> nums, int k) {
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int ans = 0;
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for (int i = 0; i < nums.size(); i++) {
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if (Integer.bitCount(i) == k) {
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ans += nums.get(i);
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}
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}
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return ans;
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}
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}
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```
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### **C++**

‎solution/2800-2899/2859.Sum of Values at Indices With K Set Bits/Solultion.java

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@@ -0,0 +1,11 @@
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class Solution {
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public int sumIndicesWithKSetBits(List<Integer> nums, int k) {
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int ans = 0;
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for (int i = 0; i < nums.size(); i++) {
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if (Integer.bitCount(i) == k) {
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ans += nums.get(i);
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}
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}
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return ans;
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}
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}

‎solution/2800-2899/2860.Happy Students/README.md

+13-1
Original file line numberDiff line numberDiff line change
@@ -71,7 +71,19 @@
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int countWays(List<Integer> nums) {
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Collections.sort(nums);
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int n = nums.size();
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int ans = 0;
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for (int i = 0; i <= n; i++) {
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if ((i == 0 || nums.get(i - 1) < i) && (i == n || nums.get(i) > i)) {
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ans++;
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}
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}
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return ans;
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}
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}
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```
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### **C++**

‎solution/2800-2899/2860.Happy Students/README_EN.md

+13-1
Original file line numberDiff line numberDiff line change
@@ -61,7 +61,19 @@ The class teacher selects all the students to form the group.
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### **Java**
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```java
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class Solution {
65+
public int countWays(List<Integer> nums) {
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Collections.sort(nums);
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int n = nums.size();
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int ans = 0;
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for (int i = 0; i <= n; i++) {
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if ((i == 0 || nums.get(i - 1) < i) && (i == n || nums.get(i) > i)) {
71+
ans++;
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}
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}
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return ans;
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}
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}
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```
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### **C++**

‎solution/2800-2899/2860.Happy Students/Solution.java

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class Solution {
2+
public int countWays(List<Integer> nums) {
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Collections.sort(nums);
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int n = nums.size();
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int ans = 0;
6+
for (int i = 0; i <= n; i++) {
7+
if ((i == 0 || nums.get(i - 1) < i) && (i == n || nums.get(i) > i)) {
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ans++;
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}
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}
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return ans;
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}
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}

‎solution/2800-2899/2861.Maximum Number of Alloys/README.md

+44-1
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@@ -94,7 +94,50 @@
9494
<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
97-
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class Solution {
98+
int n;
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int k;
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int budget;
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List<List<Integer>> composition;
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List<Integer> stock;
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List<Integer> cost;
104+
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boolean isValid(long target) {
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for (int i = 0; i < k; i++) {
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long remain = budget;
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List<Integer> currMachine = composition.get(i);
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for (int j = 0; j < n && remain >= 0; j++) {
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long need = Math.max(0, currMachine.get(j) * target - stock.get(j));
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remain -= need * cost.get(j);
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}
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if (remain >= 0) {
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return true;
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}
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}
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return false;
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}
119+
120+
public int maxNumberOfAlloys(int n, int k, int budget, List<List<Integer>> composition,
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List<Integer> stock, List<Integer> cost) {
122+
this.n = n;
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this.k = k;
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this.budget = budget;
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this.composition = composition;
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this.stock = stock;
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this.cost = cost;
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int l = -1;
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int r = budget / cost.get(0) + stock.get(0);
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while (l < r) {
131+
int mid = (l + r + 1) >> 1;
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if (isValid(mid)) {
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l = mid;
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} else {
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r = mid - 1;
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}
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}
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return l;
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}
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}
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```
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### **C++**

‎solution/2800-2899/2861.Maximum Number of Alloys/README_EN.md

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@@ -84,7 +84,50 @@ It can be proven that we can create at most 2 alloys.
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### **Java**
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```java
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class Solution {
88+
int n;
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int k;
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int budget;
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List<List<Integer>> composition;
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List<Integer> stock;
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List<Integer> cost;
94+
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boolean isValid(long target) {
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for (int i = 0; i < k; i++) {
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long remain = budget;
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List<Integer> currMachine = composition.get(i);
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for (int j = 0; j < n && remain >= 0; j++) {
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long need = Math.max(0, currMachine.get(j) * target - stock.get(j));
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remain -= need * cost.get(j);
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}
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if (remain >= 0) {
104+
return true;
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}
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}
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return false;
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}
109+
110+
public int maxNumberOfAlloys(int n, int k, int budget, List<List<Integer>> composition,
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List<Integer> stock, List<Integer> cost) {
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this.n = n;
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this.k = k;
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this.budget = budget;
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this.composition = composition;
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this.stock = stock;
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this.cost = cost;
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int l = -1;
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int r = budget / cost.get(0) + stock.get(0);
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while (l < r) {
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int mid = (l + r + 1) >> 1;
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if (isValid(mid)) {
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l = mid;
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} else {
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r = mid - 1;
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}
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}
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return l;
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}
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}
88131
```
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90133
### **C++**

‎solution/2800-2899/2861.Maximum Number of Alloys/Solution.java

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@@ -0,0 +1,44 @@
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class Solution {
2+
int n;
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int k;
4+
int budget;
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List<List<Integer>> composition;
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List<Integer> stock;
7+
List<Integer> cost;
8+
9+
boolean isValid(long target) {
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for (int i = 0; i < k; i++) {
11+
long remain = budget;
12+
List<Integer> currMachine = composition.get(i);
13+
for (int j = 0; j < n && remain >= 0; j++) {
14+
long need = Math.max(0, currMachine.get(j) * target - stock.get(j));
15+
remain -= need * cost.get(j);
16+
}
17+
if (remain >= 0) {
18+
return true;
19+
}
20+
}
21+
return false;
22+
}
23+
24+
public int maxNumberOfAlloys(int n, int k, int budget, List<List<Integer>> composition,
25+
List<Integer> stock, List<Integer> cost) {
26+
this.n = n;
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this.k = k;
28+
this.budget = budget;
29+
this.composition = composition;
30+
this.stock = stock;
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this.cost = cost;
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int l = -1;
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int r = budget / cost.get(0) + stock.get(0);
34+
while (l < r) {
35+
int mid = (l + r + 1) >> 1;
36+
if (isValid(mid)) {
37+
l = mid;
38+
} else {
39+
r = mid - 1;
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}
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}
42+
return l;
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}
44+
}

‎solution/2800-2899/2862.Maximum Element-Sum of a Complete Subset of Indices/README.md

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@@ -71,6 +71,29 @@
7171
<!-- 这里可写当前语言的特殊实现逻辑 -->
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7373
```java
74+
class Solution {
75+
public long maximumSum(List<Integer> nums) {
76+
long ans = 0;
77+
int n = nums.size();
78+
boolean[] used = new boolean[n + 1];
79+
int bound = (int) Math.floor(Math.sqrt(n));
80+
int[] squares = new int[bound + 1];
81+
for (int i = 1; i <= bound + 1; i++) {
82+
squares[i - 1] = i * i;
83+
}
84+
for (int i = 1; i <= n; i++) {
85+
long res = 0;
86+
int idx = 0;
87+
int curr = i * squares[idx];
88+
while (curr <= n) {
89+
res += nums.get(curr - 1);
90+
curr = i * squares[++idx];
91+
}
92+
ans = Math.max(ans, res);
93+
}
94+
return ans;
95+
}
96+
}
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7598
```
7699

‎solution/2800-2899/2862.Maximum Element-Sum of a Complete Subset of Indices/README_EN.md

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@@ -61,6 +61,29 @@ Hence, the maximum element-sum of a complete subset of indices is 19.
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### **Java**
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6363
```java
64+
class Solution {
65+
public long maximumSum(List<Integer> nums) {
66+
long ans = 0;
67+
int n = nums.size();
68+
boolean[] used = new boolean[n + 1];
69+
int bound = (int) Math.floor(Math.sqrt(n));
70+
int[] squares = new int[bound + 1];
71+
for (int i = 1; i <= bound + 1; i++) {
72+
squares[i - 1] = i * i;
73+
}
74+
for (int i = 1; i <= n; i++) {
75+
long res = 0;
76+
int idx = 0;
77+
int curr = i * squares[idx];
78+
while (curr <= n) {
79+
res += nums.get(curr - 1);
80+
curr = i * squares[++idx];
81+
}
82+
ans = Math.max(ans, res);
83+
}
84+
return ans;
85+
}
86+
}
6487

6588
```
6689

‎solution/2800-2899/2862.Maximum Element-Sum of a Complete Subset of Indices/Solution.java

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1+
class Solution {
2+
public long maximumSum(List<Integer> nums) {
3+
long ans = 0;
4+
int n = nums.size();
5+
boolean[] used = new boolean[n + 1];
6+
int bound = (int) Math.floor(Math.sqrt(n));
7+
int[] squares = new int[bound + 1];
8+
for (int i = 1; i <= bound + 1; i++) {
9+
squares[i - 1] = i * i;
10+
}
11+
for (int i = 1; i <= n; i++) {
12+
long res = 0;
13+
int idx = 0;
14+
int curr = i * squares[idx];
15+
while (curr <= n) {
16+
res += nums.get(curr - 1);
17+
curr = i * squares[++idx];
18+
}
19+
ans = Math.max(ans, res);
20+
}
21+
return ans;
22+
}
23+
}

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