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Commit cd2f210

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Added tasks 3179-3181
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package g3101_3200.s3179_find_the_n_th_value_after_k_seconds;
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// #Medium #Array #Math #Simulation #Prefix_Sum #Combinatorics
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// #2024_06_14_Time_2_ms_(99.86%)_Space_40.9_MB_(85.18%)
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public class Solution {
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private final int mod = (int) (Math.pow(10, 9) + 7);
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public int valueAfterKSeconds(int n, int k) {
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if (n == 1) {
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return 1;
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}
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return combination(k + n - 1, k);
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}
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private int combination(int a, int b) {
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long numerator = 1;
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long denominator = 1;
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for (int i = 0; i < b; i++) {
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numerator = (numerator * (a - i)) % mod;
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denominator = (denominator * (i + 1)) % mod;
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}
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// Calculate the modular inverse of denominator
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long denominatorInverse = power(denominator, mod - 2);
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return (int) ((numerator * denominatorInverse) % mod);
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}
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// Function to calculate power
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private long power(long x, int y) {
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long result = 1;
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while (y > 0) {
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if (y % 2 == 1) {
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result = (result * x) % mod;
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}
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y = y >> 1;
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x = (x * x) % mod;
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}
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return result;
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}
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}
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3179\. Find the N-th Value After K Seconds
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Medium
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You are given two integers `n` and `k`.
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Initially, you start with an array `a` of `n` integers where `a[i] = 1` for all `0 <= i <= n - 1`. After each second, you simultaneously update each element to be the sum of all its preceding elements plus the element itself. For example, after one second, `a[0]` remains the same, `a[1]` becomes `a[0] + a[1]`, `a[2]` becomes `a[0] + a[1] + a[2]`, and so on.
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Return the **value** of `a[n - 1]` after `k` seconds.
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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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**Example 1:**
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**Input:** n = 4, k = 5
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**Output:** 56
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**Explanation:**
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| Second | State After |
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|--------|-------------------|
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| 0 | `[1, 1, 1, 1]` |
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| 1 | `[1, 2, 3, 4]` |
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| 2 | `[1, 3, 6, 10]` |
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| 3 | `[1, 4, 10, 20]` |
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| 4 | `[1, 5, 15, 35]` |
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| 5 | `[1, 6, 21, 56]` |
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**Example 2:**
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**Input:** n = 5, k = 3
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**Output:** 35
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**Explanation:**
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| Second | State After |
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|--------|-------------------|
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| 0 | `[1, 1, 1, 1, 1]` |
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| 1 | `[1, 2, 3, 4, 5]` |
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| 2 | `[1, 3, 6, 10, 15]` |
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| 3 | `[1, 4, 10, 20, 35]` |
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**Constraints:**
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* `1 <= n, k <= 1000`
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package g3101_3200.s3180_maximum_total_reward_using_operations_i;
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// #Medium #Array #Dynamic_Programming #2024_06_14_Time_1_ms_(100.00%)_Space_43.3_MB_(97.85%)
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@SuppressWarnings("java:S135")
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public class Solution {
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private int[] sortedSet(int[] values) {
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int max = 0;
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for (int x : values) {
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if (x > max) {
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max = x;
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}
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}
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boolean[] set = new boolean[max + 1];
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int n = 0;
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for (int x : values) {
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if (!set[x]) {
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set[x] = true;
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n++;
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}
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}
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int[] result = new int[n];
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for (int x = max; x > 0; x--) {
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if (set[x]) {
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result[--n] = x;
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}
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}
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return result;
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}
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public int maxTotalReward(int[] rewardValues) {
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rewardValues = sortedSet(rewardValues);
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int n = rewardValues.length;
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int max = rewardValues[n - 1];
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boolean[] isSumPossible = new boolean[max];
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isSumPossible[0] = true;
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int maxSum = 0;
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int last = 1;
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for (int sum = rewardValues[0]; sum < max; sum++) {
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while (last < n && rewardValues[last] <= sum) {
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last++;
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}
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int s2 = sum / 2;
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for (int i = last - 1; i >= 0; i--) {
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int x = rewardValues[i];
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if (x <= s2) {
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break;
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}
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if (isSumPossible[sum - x]) {
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isSumPossible[sum] = true;
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maxSum = sum;
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break;
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}
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}
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}
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return maxSum + max;
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}
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}
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3180\. Maximum Total Reward Using Operations I
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Medium
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You are given an integer array `rewardValues` of length `n`, representing the values of rewards.
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Initially, your total reward `x` is 0, and all indices are **unmarked**. You are allowed to perform the following operation **any** number of times:
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* Choose an **unmarked** index `i` from the range `[0, n - 1]`.
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* If `rewardValues[i]` is **greater** than your current total reward `x`, then add `rewardValues[i]` to `x` (i.e., `x = x + rewardValues[i]`), and **mark** the index `i`.
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Return an integer denoting the **maximum** _total reward_ you can collect by performing the operations optimally.
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**Example 1:**
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**Input:** rewardValues = [1,1,3,3]
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**Output:** 4
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**Explanation:**
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During the operations, we can choose to mark the indices 0 and 2 in order, and the total reward will be 4, which is the maximum.
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**Example 2:**
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**Input:** rewardValues = [1,6,4,3,2]
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**Output:** 11
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**Explanation:**
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Mark the indices 0, 2, and 1 in order. The total reward will then be 11, which is the maximum.
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**Constraints:**
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* `1 <= rewardValues.length <= 2000`
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* `1 <= rewardValues[i] <= 2000`
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package g3101_3200.s3181_maximum_total_reward_using_operations_ii;
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// #Hard #Array #Dynamic_Programming #Bit_Manipulation
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// #2024_06_14_Time_2_ms_(100.00%)_Space_53.3_MB_(90.35%)
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public class Solution {
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public int maxTotalReward(int[] rewardValues) {
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int max = rewardValues[0];
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int n = 0;
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for (int i = 1; i < rewardValues.length; i++) {
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max = Math.max(max, rewardValues[i]);
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}
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boolean[] vis = new boolean[max + 1];
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for (int i : rewardValues) {
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if (!vis[i]) {
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n++;
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vis[i] = true;
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}
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}
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int[] rew = new int[n];
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int j = 0;
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for (int i = 0; i <= max; i++) {
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if (vis[i]) {
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rew[j++] = i;
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}
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}
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return rew[n - 1] + getAns(rew, n - 1, rew[n - 1] - 1);
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}
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private int getAns(int[] rewards, int i, int validLimit) {
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int res = 0;
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int j = nextElemWithinLimits(rewards, i - 1, validLimit);
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for (; j >= 0; j--) {
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if (res >= rewards[j] + Math.min(validLimit - rewards[j], rewards[j] - 1)) {
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break;
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}
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res =
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Math.max(
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res,
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rewards[j]
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+ getAns(
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rewards,
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j,
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Math.min(validLimit - rewards[j], rewards[j] - 1)));
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}
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return res;
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}
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private int nextElemWithinLimits(int[] rewards, int h, int k) {
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int l = 0;
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int resInd = -1;
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while (l <= h) {
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int m = (l + h) / 2;
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if (rewards[m] <= k) {
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resInd = m;
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l = m + 1;
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} else {
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h = m - 1;
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}
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}
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return resInd;
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}
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}
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3181\. Maximum Total Reward Using Operations II
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Hard
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You are given an integer array `rewardValues` of length `n`, representing the values of rewards.
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Initially, your total reward `x` is 0, and all indices are **unmarked**. You are allowed to perform the following operation **any** number of times:
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* Choose an **unmarked** index `i` from the range `[0, n - 1]`.
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* If `rewardValues[i]` is **greater** than your current total reward `x`, then add `rewardValues[i]` to `x` (i.e., `x = x + rewardValues[i]`), and **mark** the index `i`.
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Return an integer denoting the **maximum** _total reward_ you can collect by performing the operations optimally.
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**Example 1:**
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**Input:** rewardValues = [1,1,3,3]
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**Output:** 4
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**Explanation:**
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During the operations, we can choose to mark the indices 0 and 2 in order, and the total reward will be 4, which is the maximum.
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**Example 2:**
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**Input:** rewardValues = [1,6,4,3,2]
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**Output:** 11
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**Explanation:**
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Mark the indices 0, 2, and 1 in order. The total reward will then be 11, which is the maximum.
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**Constraints:**
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* <code>1 <= rewardValues.length <= 5 * 10<sup>4</sup></code>
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* <code>1 <= rewardValues[i] <= 5 * 10<sup>4</sup></code>
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package g3101_3200.s3179_find_the_n_th_value_after_k_seconds;
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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 valueAfterKSeconds() {
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assertThat(new Solution().valueAfterKSeconds(4, 5), equalTo(56));
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}
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@Test
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void valueAfterKSeconds2() {
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assertThat(new Solution().valueAfterKSeconds(5, 3), equalTo(35));
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}
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}
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package g3101_3200.s3180_maximum_total_reward_using_operations_i;
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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 maxTotalReward() {
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assertThat(new Solution().maxTotalReward(new int[] {1, 1, 3, 3}), equalTo(4));
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}
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@Test
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void maxTotalReward2() {
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assertThat(new Solution().maxTotalReward(new int[] {1, 6, 4, 3, 2}), equalTo(11));
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}
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}
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package g3101_3200.s3181_maximum_total_reward_using_operations_ii;
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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 maxTotalReward() {
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assertThat(new Solution().maxTotalReward(new int[] {1, 1, 3, 3}), equalTo(4));
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}
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@Test
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void maxTotalReward2() {
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assertThat(new Solution().maxTotalReward(new int[] {1, 6, 4, 3, 2}), equalTo(11));
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}
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}

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