diff --git a/src/main/java/g3101_3200/s3179_find_the_n_th_value_after_k_seconds/Solution.java b/src/main/java/g3101_3200/s3179_find_the_n_th_value_after_k_seconds/Solution.java
new file mode 100644
index 000000000..52978dc1f
--- /dev/null
+++ b/src/main/java/g3101_3200/s3179_find_the_n_th_value_after_k_seconds/Solution.java
@@ -0,0 +1,40 @@
+package g3101_3200.s3179_find_the_n_th_value_after_k_seconds;
+
+// #Medium #Array #Math #Simulation #Prefix_Sum #Combinatorics
+// #2024_06_14_Time_2_ms_(99.86%)_Space_40.9_MB_(85.18%)
+
+public class Solution {
+    private final int mod = (int) (Math.pow(10, 9) + 7);
+
+    public int valueAfterKSeconds(int n, int k) {
+        if (n == 1) {
+            return 1;
+        }
+        return combination(k + n - 1, k);
+    }
+
+    private int combination(int a, int b) {
+        long numerator = 1;
+        long denominator = 1;
+        for (int i = 0; i < b; i++) {
+            numerator = (numerator * (a - i)) % mod;
+            denominator = (denominator * (i + 1)) % mod;
+        }
+        // Calculate the modular inverse of denominator
+        long denominatorInverse = power(denominator, mod - 2);
+        return (int) ((numerator * denominatorInverse) % mod);
+    }
+
+    // Function to calculate power
+    private long power(long x, int y) {
+        long result = 1;
+        while (y > 0) {
+            if (y % 2 == 1) {
+                result = (result * x) % mod;
+            }
+            y = y >> 1;
+            x = (x * x) % mod;
+        }
+        return result;
+    }
+}
diff --git a/src/main/java/g3101_3200/s3179_find_the_n_th_value_after_k_seconds/readme.md b/src/main/java/g3101_3200/s3179_find_the_n_th_value_after_k_seconds/readme.md
new file mode 100644
index 000000000..5c0f8c7c0
--- /dev/null
+++ b/src/main/java/g3101_3200/s3179_find_the_n_th_value_after_k_seconds/readme.md
@@ -0,0 +1,47 @@
+3179\. Find the N-th Value After K Seconds
+
+Medium
+
+You are given two integers `n` and `k`.
+
+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.
+
+Return the **value** of `a[n - 1]` after `k` seconds.
+
+Since the answer may be very large, return it **modulo** <code>10<sup>9</sup> + 7</code>.
+
+**Example 1:**
+
+**Input:** n = 4, k = 5
+
+**Output:** 56
+
+**Explanation:**
+
+| Second | State After      |
+|--------|-------------------|
+| 0      | `[1, 1, 1, 1]`   |
+| 1      | `[1, 2, 3, 4]`   |
+| 2      | `[1, 3, 6, 10]`  |
+| 3      | `[1, 4, 10, 20]` |
+| 4      | `[1, 5, 15, 35]` |
+| 5      | `[1, 6, 21, 56]` |
+
+**Example 2:**
+
+**Input:** n = 5, k = 3
+
+**Output:** 35
+
+**Explanation:**
+
+| Second | State After       |
+|--------|-------------------|
+| 0      | `[1, 1, 1, 1, 1]` |
+| 1      | `[1, 2, 3, 4, 5]` |
+| 2      | `[1, 3, 6, 10, 15]` |
+| 3      | `[1, 4, 10, 20, 35]` |
+
+**Constraints:**
+
+*   `1 <= n, k <= 1000`
\ No newline at end of file
diff --git a/src/main/java/g3101_3200/s3180_maximum_total_reward_using_operations_i/Solution.java b/src/main/java/g3101_3200/s3180_maximum_total_reward_using_operations_i/Solution.java
new file mode 100644
index 000000000..75a33fc36
--- /dev/null
+++ b/src/main/java/g3101_3200/s3180_maximum_total_reward_using_operations_i/Solution.java
@@ -0,0 +1,58 @@
+package g3101_3200.s3180_maximum_total_reward_using_operations_i;
+
+// #Medium #Array #Dynamic_Programming #2024_06_14_Time_1_ms_(100.00%)_Space_43.3_MB_(97.85%)
+
+@SuppressWarnings("java:S135")
+public class Solution {
+    private int[] sortedSet(int[] values) {
+        int max = 0;
+        for (int x : values) {
+            if (x > max) {
+                max = x;
+            }
+        }
+        boolean[] set = new boolean[max + 1];
+        int n = 0;
+        for (int x : values) {
+            if (!set[x]) {
+                set[x] = true;
+                n++;
+            }
+        }
+        int[] result = new int[n];
+        for (int x = max; x > 0; x--) {
+            if (set[x]) {
+                result[--n] = x;
+            }
+        }
+        return result;
+    }
+
+    public int maxTotalReward(int[] rewardValues) {
+        rewardValues = sortedSet(rewardValues);
+        int n = rewardValues.length;
+        int max = rewardValues[n - 1];
+        boolean[] isSumPossible = new boolean[max];
+        isSumPossible[0] = true;
+        int maxSum = 0;
+        int last = 1;
+        for (int sum = rewardValues[0]; sum < max; sum++) {
+            while (last < n && rewardValues[last] <= sum) {
+                last++;
+            }
+            int s2 = sum / 2;
+            for (int i = last - 1; i >= 0; i--) {
+                int x = rewardValues[i];
+                if (x <= s2) {
+                    break;
+                }
+                if (isSumPossible[sum - x]) {
+                    isSumPossible[sum] = true;
+                    maxSum = sum;
+                    break;
+                }
+            }
+        }
+        return maxSum + max;
+    }
+}
diff --git a/src/main/java/g3101_3200/s3180_maximum_total_reward_using_operations_i/readme.md b/src/main/java/g3101_3200/s3180_maximum_total_reward_using_operations_i/readme.md
new file mode 100644
index 000000000..1e99f1b4b
--- /dev/null
+++ b/src/main/java/g3101_3200/s3180_maximum_total_reward_using_operations_i/readme.md
@@ -0,0 +1,37 @@
+3180\. Maximum Total Reward Using Operations I
+
+Medium
+
+You are given an integer array `rewardValues` of length `n`, representing the values of rewards.
+
+Initially, your total reward `x` is 0, and all indices are **unmarked**. You are allowed to perform the following operation **any** number of times:
+
+*   Choose an **unmarked** index `i` from the range `[0, n - 1]`.
+*   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`.
+
+Return an integer denoting the **maximum** _total reward_ you can collect by performing the operations optimally.
+
+**Example 1:**
+
+**Input:** rewardValues = [1,1,3,3]
+
+**Output:** 4
+
+**Explanation:**
+
+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.
+
+**Example 2:**
+
+**Input:** rewardValues = [1,6,4,3,2]
+
+**Output:** 11
+
+**Explanation:**
+
+Mark the indices 0, 2, and 1 in order. The total reward will then be 11, which is the maximum.
+
+**Constraints:**
+
+*   `1 <= rewardValues.length <= 2000`
+*   `1 <= rewardValues[i] <= 2000`
\ No newline at end of file
diff --git a/src/main/java/g3101_3200/s3181_maximum_total_reward_using_operations_ii/Solution.java b/src/main/java/g3101_3200/s3181_maximum_total_reward_using_operations_ii/Solution.java
new file mode 100644
index 000000000..0c6dea269
--- /dev/null
+++ b/src/main/java/g3101_3200/s3181_maximum_total_reward_using_operations_ii/Solution.java
@@ -0,0 +1,63 @@
+package g3101_3200.s3181_maximum_total_reward_using_operations_ii;
+
+// #Hard #Array #Dynamic_Programming #Bit_Manipulation
+// #2024_06_14_Time_2_ms_(100.00%)_Space_53.3_MB_(90.35%)
+
+public class Solution {
+    public int maxTotalReward(int[] rewardValues) {
+        int max = rewardValues[0];
+        int n = 0;
+        for (int i = 1; i < rewardValues.length; i++) {
+            max = Math.max(max, rewardValues[i]);
+        }
+        boolean[] vis = new boolean[max + 1];
+        for (int i : rewardValues) {
+            if (!vis[i]) {
+                n++;
+                vis[i] = true;
+            }
+        }
+        int[] rew = new int[n];
+        int j = 0;
+        for (int i = 0; i <= max; i++) {
+            if (vis[i]) {
+                rew[j++] = i;
+            }
+        }
+        return rew[n - 1] + getAns(rew, n - 1, rew[n - 1] - 1);
+    }
+
+    private int getAns(int[] rewards, int i, int validLimit) {
+        int res = 0;
+        int j = nextElemWithinLimits(rewards, i - 1, validLimit);
+        for (; j >= 0; j--) {
+            if (res >= rewards[j] + Math.min(validLimit - rewards[j], rewards[j] - 1)) {
+                break;
+            }
+            res =
+                    Math.max(
+                            res,
+                            rewards[j]
+                                    + getAns(
+                                            rewards,
+                                            j,
+                                            Math.min(validLimit - rewards[j], rewards[j] - 1)));
+        }
+        return res;
+    }
+
+    private int nextElemWithinLimits(int[] rewards, int h, int k) {
+        int l = 0;
+        int resInd = -1;
+        while (l <= h) {
+            int m = (l + h) / 2;
+            if (rewards[m] <= k) {
+                resInd = m;
+                l = m + 1;
+            } else {
+                h = m - 1;
+            }
+        }
+        return resInd;
+    }
+}
diff --git a/src/main/java/g3101_3200/s3181_maximum_total_reward_using_operations_ii/readme.md b/src/main/java/g3101_3200/s3181_maximum_total_reward_using_operations_ii/readme.md
new file mode 100644
index 000000000..8f40bfbc0
--- /dev/null
+++ b/src/main/java/g3101_3200/s3181_maximum_total_reward_using_operations_ii/readme.md
@@ -0,0 +1,37 @@
+3181\. Maximum Total Reward Using Operations II
+
+Hard
+
+You are given an integer array `rewardValues` of length `n`, representing the values of rewards.
+
+Initially, your total reward `x` is 0, and all indices are **unmarked**. You are allowed to perform the following operation **any** number of times:
+
+*   Choose an **unmarked** index `i` from the range `[0, n - 1]`.
+*   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`.
+
+Return an integer denoting the **maximum** _total reward_ you can collect by performing the operations optimally.
+
+**Example 1:**
+
+**Input:** rewardValues = [1,1,3,3]
+
+**Output:** 4
+
+**Explanation:**
+
+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.
+
+**Example 2:**
+
+**Input:** rewardValues = [1,6,4,3,2]
+
+**Output:** 11
+
+**Explanation:**
+
+Mark the indices 0, 2, and 1 in order. The total reward will then be 11, which is the maximum.
+
+**Constraints:**
+
+*   <code>1 <= rewardValues.length <= 5 * 10<sup>4</sup></code>
+*   <code>1 <= rewardValues[i] <= 5 * 10<sup>4</sup></code>
\ No newline at end of file
diff --git a/src/test/java/g3101_3200/s3179_find_the_n_th_value_after_k_seconds/SolutionTest.java b/src/test/java/g3101_3200/s3179_find_the_n_th_value_after_k_seconds/SolutionTest.java
new file mode 100644
index 000000000..6dfb316f7
--- /dev/null
+++ b/src/test/java/g3101_3200/s3179_find_the_n_th_value_after_k_seconds/SolutionTest.java
@@ -0,0 +1,18 @@
+package g3101_3200.s3179_find_the_n_th_value_after_k_seconds;
+
+import static org.hamcrest.CoreMatchers.equalTo;
+import static org.hamcrest.MatcherAssert.assertThat;
+
+import org.junit.jupiter.api.Test;
+
+class SolutionTest {
+    @Test
+    void valueAfterKSeconds() {
+        assertThat(new Solution().valueAfterKSeconds(4, 5), equalTo(56));
+    }
+
+    @Test
+    void valueAfterKSeconds2() {
+        assertThat(new Solution().valueAfterKSeconds(5, 3), equalTo(35));
+    }
+}
diff --git a/src/test/java/g3101_3200/s3180_maximum_total_reward_using_operations_i/SolutionTest.java b/src/test/java/g3101_3200/s3180_maximum_total_reward_using_operations_i/SolutionTest.java
new file mode 100644
index 000000000..8473316e3
--- /dev/null
+++ b/src/test/java/g3101_3200/s3180_maximum_total_reward_using_operations_i/SolutionTest.java
@@ -0,0 +1,18 @@
+package g3101_3200.s3180_maximum_total_reward_using_operations_i;
+
+import static org.hamcrest.CoreMatchers.equalTo;
+import static org.hamcrest.MatcherAssert.assertThat;
+
+import org.junit.jupiter.api.Test;
+
+class SolutionTest {
+    @Test
+    void maxTotalReward() {
+        assertThat(new Solution().maxTotalReward(new int[] {1, 1, 3, 3}), equalTo(4));
+    }
+
+    @Test
+    void maxTotalReward2() {
+        assertThat(new Solution().maxTotalReward(new int[] {1, 6, 4, 3, 2}), equalTo(11));
+    }
+}
diff --git a/src/test/java/g3101_3200/s3181_maximum_total_reward_using_operations_ii/SolutionTest.java b/src/test/java/g3101_3200/s3181_maximum_total_reward_using_operations_ii/SolutionTest.java
new file mode 100644
index 000000000..5457bb618
--- /dev/null
+++ b/src/test/java/g3101_3200/s3181_maximum_total_reward_using_operations_ii/SolutionTest.java
@@ -0,0 +1,18 @@
+package g3101_3200.s3181_maximum_total_reward_using_operations_ii;
+
+import static org.hamcrest.CoreMatchers.equalTo;
+import static org.hamcrest.MatcherAssert.assertThat;
+
+import org.junit.jupiter.api.Test;
+
+class SolutionTest {
+    @Test
+    void maxTotalReward() {
+        assertThat(new Solution().maxTotalReward(new int[] {1, 1, 3, 3}), equalTo(4));
+    }
+
+    @Test
+    void maxTotalReward2() {
+        assertThat(new Solution().maxTotalReward(new int[] {1, 6, 4, 3, 2}), equalTo(11));
+    }
+}