diff --git a/solution/0600-0699/0698.Partition to K Equal Sum Subsets/README.md b/solution/0600-0699/0698.Partition to K Equal Sum Subsets/README.md
index b259dac9984f0..ba4e8344bd56c 100644
--- a/solution/0600-0699/0698.Partition to K Equal Sum Subsets/README.md	
+++ b/solution/0600-0699/0698.Partition to K Equal Sum Subsets/README.md	
@@ -56,13 +56,13 @@ tags:
 
 ### 方法一：DFS + 剪枝
 
-根据题意，我们需要将数组 `nums` 划分为 $k$ 个子集，且每个子集的和相等。因此，先累加 `nums` 中所有元素的和，如果不能被 $k$ 整除，说明无法划分为 $k$ 个子集，提前返回 `false`。
+根据题意，我们需要将数组 $\textit{nums}$ 划分为 $k$ 个子集，且每个子集的和相等。因此，先累加 $\textit{nums}$ 中所有元素的和，如果不能被 $k$ 整除，说明无法划分为 $k$ 个子集，提前返回 $\textit{false}$。
 
-如果能被 $k$ 整除，不妨将每个子集期望的和记为 $s$，然后创建一个长度为 $k$ 的数组 `cur`，表示当前每个子集的和。
+如果能被 $k$ 整除，不妨将每个子集期望的和记为 $s$，然后创建一个长度为 $k$ 的数组 $\textit{cur}$，表示当前每个子集的和。
 
-对数组 `nums` 进行降序排序（减少搜索次数），然后从第一个元素开始，依次尝试将其加入到 `cur` 的每个子集中。这里如果将 `nums[i]` 加入某个子集 `cur[j]` 后，子集的和超过 $s$，说明无法放入，可以直接跳过；另外，如果 `cur[j]` 与 `cur[j - 1]` 相等，意味着我们在 `cur[j - 1]` 的时候已经完成了搜索，也可以跳过当前的搜索。
+对数组 $\textit{nums}$ 进行降序排序（减少搜索次数），然后从第一个元素开始，依次尝试将其加入到 $\textit{cur}$ 的每个子集中。这里如果将 $\textit{nums}[i]$ 加入某个子集 $\textit{cur}[j]$ 后，子集的和超过 $s$，说明无法放入，可以直接跳过；另外，如果 $\textit{cur}[j]$ 与 $\textit{cur}[j - 1]$ 相等，意味着我们在 $\textit{cur}[j - 1]$ 的时候已经完成了搜索，也可以跳过当前的搜索。
 
-如果能将所有元素都加入到 `cur` 中，说明可以划分为 $k$ 个子集，返回 `true`。
+如果能将所有元素都加入到 $\textit{cur}$ 中，说明可以划分为 $k$ 个子集，返回 $\textit{true}$。
 
 <!-- tabs:start -->
 
@@ -145,8 +145,7 @@ public:
         s /= k;
         int n = nums.size();
         vector<int> cur(k);
-        function<bool(int)> dfs;
-        dfs = [&](int i) {
+        function<bool(int)> dfs = [&](int i) {
             if (i == n) {
                 return true;
             }
@@ -210,31 +209,32 @@ func canPartitionKSubsets(nums []int, k int) bool {
 
 ```ts
 function canPartitionKSubsets(nums: number[], k: number): boolean {
-    let s = nums.reduce((a, b) => a + b);
-    if (s % k !== 0) {
-        return false;
-    }
-    s /= k;
-    nums.sort((a, b) => a - b);
-    const n = nums.length;
-    const f: boolean[] = new Array(1 << n).fill(false);
-    f[0] = true;
-    const cur: number[] = new Array(n).fill(0);
-    for (let i = 0; i < 1 << n; ++i) {
-        if (!f[i]) {
-            continue;
+    const dfs = (i: number): boolean => {
+        if (i === nums.length) {
+            return true;
         }
-        for (let j = 0; j < n; ++j) {
-            if (cur[i] + nums[j] > s) {
-                break;
+        for (let j = 0; j < k; j++) {
+            if (j > 0 && cur[j] === cur[j - 1]) {
+                continue;
             }
-            if (((i >> j) & 1) === 0) {
-                f[i | (1 << j)] = true;
-                cur[i | (1 << j)] = (cur[i] + nums[j]) % s;
+            cur[j] += nums[i];
+            if (cur[j] <= s && dfs(i + 1)) {
+                return true;
             }
+            cur[j] -= nums[i];
         }
+        return false;
+    };
+
+    let s = nums.reduce((a, b) => a + b, 0);
+    const mod = s % k;
+    if (mod !== 0) {
+        return false;
     }
-    return f[(1 << n) - 1];
+    s = Math.floor(s / k);
+    const cur = Array(k).fill(0);
+    nums.sort((a, b) => b - a);
+    return dfs(0);
 }
 ```
 
@@ -246,22 +246,22 @@ function canPartitionKSubsets(nums: number[], k: number): boolean {
 
 ### 方法二：状态压缩 + 记忆化搜索
 
-与方法一相同，我们依然先判断数组 `nums` 是否有可能被划分为 $k$ 个子集。如果不能被 $k$ 整除，直接返回 `false`。
+与方法一相同，我们依然先判断数组 $\textit{nums}$ 是否有可能被划分为 $k$ 个子集。如果不能被 $k$ 整除，直接返回 $\textit{false}$。
 
-我们记 $s$ 为每个子集期望的和，当前元素被划分的情况为 `state`。对于第 $i$ 个数，若 `((state >> i) & 1)` 等于 $0$，说明第 $i$ 个元素未被划分。
+我们记 $s$ 为每个子集期望的和，当前元素被划分的情况为 $\textit{state}$。对于第 $i$ 个数，若 $\textit{state}$ 的第 $i$ 位为 $0$，说明第 $i$ 个元素未被划分。
 
 我们的目标是从全部元素中凑出 $k$ 个和为 $s$ 的子集。记当前子集的和为 $t$。在未划分第 $i$ 个元素时：
 
--   若 $t + nums[i] \gt s$，说明第 $i$ 个元素不能被添加到当前子集中，由于我们对 `nums` 数组进行升序排列，因此数组 `nums` 从位置 $i$ 开始的所有元素都不能被添加到当前子集，直接返回 `false`。
--   否则，将第 $i$ 个元素添加到当前子集中，状态变为 `state | (1 << i)`，然后继续对未划分的元素进行搜索。需要注意的是，若 $t + nums[i] = s$，说明恰好可以得到一个和为 $s$ 的子集，下一步将 $t$ 归零（可以通过 `(t + nums[i]) % s` 实现），并继续划分下一个子集。
+-   若 $t + \textit{nums}[i] \gt s$，说明第 $i$ 个元素不能被添加到当前子集中，由于我们对 $\textit{nums}$ 数组进行升序排列，因此数组 $\textit{nums}$ 从位置 $i$ 开始的所有元素都不能被添加到当前子集，直接返回 $\textit{false}$。
+-   否则，将第 $i$ 个元素添加到当前子集中，状态变为 $\textit{state} | 2^i$，然后继续对未划分的元素进行搜索。需要注意的是，若 $t + \textit{nums}[i] = s$，说明恰好可以得到一个和为 $s$ 的子集，下一步将 $t$ 归零（可以通过 $(t + \textit{nums}[i]) \bmod s$ 实现），并继续划分下一个子集。
 
-为了避免重复搜索，我们使用一个长度为 $2^n$ 的数组 `f` 记录每个状态下的搜索结果。数组 `f` 有三个可能的值：
+为了避免重复搜索，我们使用一个长度为 $2^n$ 的数组 $\textit{f}$ 记录每个状态下的搜索结果。数组 $\textit{f}$ 有三个可能的值：
 
 -   `0`：表示当前状态还未搜索过；
 -   `-1`：表示当前状态下无法划分为 $k$ 个子集；
 -   `1`：表示当前状态下可以划分为 $k$ 个子集。
 
-时间复杂度 $O(n\times 2^n)$，空间复杂度 $O(2^n)$。其中 $n$ 表示数组 $nums$ 的长度。对于每个状态，我们需要遍历数组 `nums`，时间复杂度为 $O(n)$；状态总数为 $2^n$，因此总的时间复杂度为 $O(n\times 2^n)$。
+时间复杂度 $O(n \times 2^n)$，空间复杂度 $O(2^n)$。其中 $n$ 表示数组 $\textit{nums}$ 的长度。对于每个状态，我们需要遍历数组 $\textit{nums}$，时间复杂度为 $O(n)$；状态总数为 $2^n$，因此总的时间复杂度为 $O(n\times 2^n)$。
 
 <!-- tabs:start -->
 
@@ -355,8 +355,7 @@ public:
         int n = nums.size();
         int mask = (1 << n) - 1;
         vector<int> f(1 << n);
-        function<bool(int, int)> dfs;
-        dfs = [&](int state, int t) {
+        function<bool(int, int)> dfs = [&](int state, int t) {
             if (state == mask) {
                 return true;
             }
@@ -428,6 +427,47 @@ func canPartitionKSubsets(nums []int, k int) bool {
 }
 ```
 
+#### TypeScript
+
+```ts
+function canPartitionKSubsets(nums: number[], k: number): boolean {
+    let s = nums.reduce((a, b) => a + b, 0);
+    if (s % k !== 0) {
+        return false;
+    }
+    s = Math.floor(s / k);
+    nums.sort((a, b) => a - b);
+    const n = nums.length;
+    const mask = (1 << n) - 1;
+    const f = Array(1 << n).fill(0);
+
+    const dfs = (state: number, t: number): boolean => {
+        if (state === mask) {
+            return true;
+        }
+        if (f[state] !== 0) {
+            return f[state] === 1;
+        }
+        for (let i = 0; i < n; ++i) {
+            if ((state >> i) & 1) {
+                continue;
+            }
+            if (t + nums[i] > s) {
+                break;
+            }
+            if (dfs(state | (1 << i), (t + nums[i]) % s)) {
+                f[state] = 1;
+                return true;
+            }
+        }
+        f[state] = -1;
+        return false;
+    };
+
+    return dfs(0, 0);
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->
@@ -438,13 +478,13 @@ func canPartitionKSubsets(nums []int, k int) bool {
 
 我们可以使用动态规划的方法求解本题。
 
-我们定义 $f[i]$ 表示当前选取的数字的状态为 $i$ 时，是否存在 $k$ 个子集满足题目要求。初始时 $f[0]=true$，答案为 $f[2^n-1]$。其中 $n$ 表示数组 $nums$ 的长度。另外，我们定义 $cur[i]$ 表示当前选取的数字的状态为 $i$ 时，最后一个子集的和。
+我们定义 $f[i]$ 表示当前选取的数字的状态为 $i$ 时，是否存在 $k$ 个子集满足题目要求。初始时 $f[0]= true$，答案为 $f[2^n-1]$。其中 $n$ 表示数组 $nums$ 的长度。另外，我们定义 $cur[i]$ 表示当前选取的数字的状态为 $i$ 时，最后一个子集的和。
 
-我们在 $[0,2^n)$ 的范围内枚举状态 $i$，对于每个状态 $i$，如果 $f[i]$ 为 `false`，我们直接跳过即可。否则，我们枚举 $nums$ 数组中的任意一个数 $nums[j]$，如果 $cur[i] + nums[j] \gt s$，我们直接跳出枚举循环，因为后面的数更大，无法放入当前子集；否则，如果 $i$ 的二进制表示的第 $j$ 位为 $0$，说明当前 $nums[j]$ 还没有被选取，我们可以将其放入当前子集中，此时状态变为 $i | 2^j$，并更新 $cur[i | 2^j] = (cur[i] + nums[j]) \bmod s$，并且 $f[i | 2^j] = true$。
+我们在 $[0, 2^n]$ 的范围内枚举状态 $i$，对于每个状态 $i$，如果 $f[i]$ 为 $\textit{false}$，我们直接跳过即可。否则，我们枚举 $\textit{nums}$ 数组中的任意一个数 $\textit{nums}[j]$，如果 $\textit{cur}[i] + \textit{nums}[j] > s$，我们直接跳出枚举循环，因为后面的数更大，无法放入当前子集；否则，如果 $i$ 的二进制表示的第 $j$ 位为 $0$，说明当前 $\textit{nums}[j]$ 还没有被选取，我们可以将其放入当前子集中，此时状态变为 $i | 2^j$，并更新 $\textit{cur}[i | 2^j] = (\textit{cur}[i] + \textit{nums}[j]) \bmod s$，并且 $f[i | 2^j] = \textit{true}$。
 
-最后，我们返回 $f[2^n-1]$ 即可。
+最后，我们返回 $f[2^n - 1]$ 即可。
 
-时间复杂度 $O(n \times 2^n)$，空间复杂度 $O(2^n)$。其中 $n$ 表示数组 $nums$ 的长度。
+时间复杂度 $O(n \times 2^n)$，空间复杂度 $O(2^n)$。其中 $n$ 表示数组 $\textit{nums}$ 的长度。
 
 <!-- tabs:start -->
 
@@ -584,6 +624,38 @@ func canPartitionKSubsets(nums []int, k int) bool {
 }
 ```
 
+#### TypeScript
+
+```ts
+function canPartitionKSubsets(nums: number[], k: number): boolean {
+    let s = nums.reduce((a, b) => a + b);
+    if (s % k !== 0) {
+        return false;
+    }
+    s /= k;
+    nums.sort((a, b) => a - b);
+    const n = nums.length;
+    const f: boolean[] = Array(1 << n).fill(false);
+    f[0] = true;
+    const cur: number[] = Array(n).fill(0);
+    for (let i = 0; i < 1 << n; ++i) {
+        if (!f[i]) {
+            continue;
+        }
+        for (let j = 0; j < n; ++j) {
+            if (cur[i] + nums[j] > s) {
+                break;
+            }
+            if (((i >> j) & 1) === 0) {
+                f[i | (1 << j)] = true;
+                cur[i | (1 << j)] = (cur[i] + nums[j]) % s;
+            }
+        }
+    }
+    return f[(1 << n) - 1];
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->
diff --git a/solution/0600-0699/0698.Partition to K Equal Sum Subsets/README_EN.md b/solution/0600-0699/0698.Partition to K Equal Sum Subsets/README_EN.md
index 0904227f64a6a..7e8cd3303f64d 100644
--- a/solution/0600-0699/0698.Partition to K Equal Sum Subsets/README_EN.md	
+++ b/solution/0600-0699/0698.Partition to K Equal Sum Subsets/README_EN.md	
@@ -54,7 +54,15 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: DFS + Pruning
+
+According to the problem description, we need to partition the array $\textit{nums}$ into $k$ subsets such that the sum of each subset is equal. Therefore, we first sum all the elements in $\textit{nums}$. If the total sum cannot be divided by $k$, it means we cannot partition the array into $k$ subsets, and we return $\textit{false}$ early.
+
+If the total sum can be divided by $k$, let's denote the expected sum of each subset as $s$. Then, we create an array $\textit{cur}$ of length $k$ to represent the current sum of each subset.
+
+We sort the array $\textit{nums}$ in descending order (to reduce the number of searches), and then start from the first element, trying to add it to each subset in $\textit{cur}$ one by one. If adding $\textit{nums}[i]$ to a subset $\textit{cur}[j]$ makes the subset's sum exceed $s$, it means it cannot be placed in that subset, and we can skip it. Additionally, if $\textit{cur}[j]$ is equal to $\textit{cur}[j - 1]$, it means we have already completed the search for $\textit{cur}[j - 1]$, and we can skip the current search.
+
+If we can add all elements to $\textit{cur}$, it means we can partition the array into $k$ subsets, and we return $\textit{true}$.
 
 <!-- tabs:start -->
 
@@ -63,7 +71,7 @@ tags:
 ```python
 class Solution:
     def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
-        def dfs(i):
+        def dfs(i: int) -> bool:
             if i == len(nums):
                 return True
             for j in range(k):
@@ -137,8 +145,7 @@ public:
         s /= k;
         int n = nums.size();
         vector<int> cur(k);
-        function<bool(int)> dfs;
-        dfs = [&](int i) {
+        function<bool(int)> dfs = [&](int i) {
             if (i == n) {
                 return true;
             }
@@ -202,31 +209,32 @@ func canPartitionKSubsets(nums []int, k int) bool {
 
 ```ts
 function canPartitionKSubsets(nums: number[], k: number): boolean {
-    let s = nums.reduce((a, b) => a + b);
-    if (s % k !== 0) {
-        return false;
-    }
-    s /= k;
-    nums.sort((a, b) => a - b);
-    const n = nums.length;
-    const f: boolean[] = new Array(1 << n).fill(false);
-    f[0] = true;
-    const cur: number[] = new Array(n).fill(0);
-    for (let i = 0; i < 1 << n; ++i) {
-        if (!f[i]) {
-            continue;
+    const dfs = (i: number): boolean => {
+        if (i === nums.length) {
+            return true;
         }
-        for (let j = 0; j < n; ++j) {
-            if (cur[i] + nums[j] > s) {
-                break;
+        for (let j = 0; j < k; j++) {
+            if (j > 0 && cur[j] === cur[j - 1]) {
+                continue;
             }
-            if (((i >> j) & 1) === 0) {
-                f[i | (1 << j)] = true;
-                cur[i | (1 << j)] = (cur[i] + nums[j]) % s;
+            cur[j] += nums[i];
+            if (cur[j] <= s && dfs(i + 1)) {
+                return true;
             }
+            cur[j] -= nums[i];
         }
+        return false;
+    };
+
+    let s = nums.reduce((a, b) => a + b, 0);
+    const mod = s % k;
+    if (mod !== 0) {
+        return false;
     }
-    return f[(1 << n) - 1];
+    s = Math.floor(s / k);
+    const cur = Array(k).fill(0);
+    nums.sort((a, b) => b - a);
+    return dfs(0);
 }
 ```
 
@@ -236,7 +244,24 @@ function canPartitionKSubsets(nums: number[], k: number): boolean {
 
 <!-- solution:start -->
 
-### Solution 2
+### Solution 2: State Compression + Memoization
+
+Similar to Solution 1, we first check whether the array $\textit{nums}$ can be partitioned into $k$ subsets. If it cannot be divided by $k$, we directly return $\textit{false}$.
+
+Let $s$ be the expected sum of each subset, and let $\textit{state}$ represent the current partitioning state of the elements. For the $i$-th number, if the $i$-th bit of $\textit{state}$ is $0$, it means the $i$-th element has not been partitioned.
+
+Our goal is to form $k$ subsets with a sum of $s$ from all elements. Let $t$ be the current sum of the subset. When the $i$-th element is not partitioned:
+
+-   If $t + \textit{nums}[i] \gt s$, it means the $i$-th element cannot be added to the current subset. Since we sort the array $\textit{nums}$ in ascending order, all elements from position $i$ onwards cannot be added to the current subset, and we directly return $\textit{false}$.
+-   Otherwise, add the $i$-th element to the current subset, change the state to $\textit{state} | 2^i$, and continue searching for unpartitioned elements. Note that if $t + \textit{nums}[i] = s$, it means we can form a subset with a sum of $s$. The next step is to reset $t$ to zero (which can be achieved by $(t + \textit{nums}[i]) \bmod s$) and continue partitioning the next subset.
+
+To avoid repeated searches, we use an array $\textit{f}$ of length $2^n$ to record the search results for each state. The array $\textit{f}$ has three possible values:
+
+-   `0` indicates that the current state has not been searched yet;
+-   `-1`: indicates that the current state cannot be partitioned into $k$ subsets;
+-   `1`: indicates that the current state can be partitioned into $k$ subsets.
+
+The time complexity is $O(n \times 2^n)$, and the space complexity is $O(2^n)$. Here, $n$ represents the length of the array $\textit{nums}$. For each state, we need to traverse the array $\textit{nums}$, which has a time complexity of $O(n)$. The total number of states is $2^n$, so the overall time complexity is $O(n \times 2^n)$.
 
 <!-- tabs:start -->
 
@@ -330,8 +355,7 @@ public:
         int n = nums.size();
         int mask = (1 << n) - 1;
         vector<int> f(1 << n);
-        function<bool(int, int)> dfs;
-        dfs = [&](int state, int t) {
+        function<bool(int, int)> dfs = [&](int state, int t) {
             if (state == mask) {
                 return true;
             }
@@ -403,13 +427,64 @@ func canPartitionKSubsets(nums []int, k int) bool {
 }
 ```
 
+#### TypeScript
+
+```ts
+function canPartitionKSubsets(nums: number[], k: number): boolean {
+    let s = nums.reduce((a, b) => a + b, 0);
+    if (s % k !== 0) {
+        return false;
+    }
+    s = Math.floor(s / k);
+    nums.sort((a, b) => a - b);
+    const n = nums.length;
+    const mask = (1 << n) - 1;
+    const f = Array(1 << n).fill(0);
+
+    const dfs = (state: number, t: number): boolean => {
+        if (state === mask) {
+            return true;
+        }
+        if (f[state] !== 0) {
+            return f[state] === 1;
+        }
+        for (let i = 0; i < n; ++i) {
+            if ((state >> i) & 1) {
+                continue;
+            }
+            if (t + nums[i] > s) {
+                break;
+            }
+            if (dfs(state | (1 << i), (t + nums[i]) % s)) {
+                f[state] = 1;
+                return true;
+            }
+        }
+        f[state] = -1;
+        return false;
+    };
+
+    return dfs(0, 0);
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->
 
 <!-- solution:start -->
 
-### Solution 3
+### Solution 3: Dynamic Programming
+
+We can use dynamic programming to solve this problem.
+
+We define $f[i]$ to represent whether there exists $k$ subsets that meet the requirements when the current selected numbers' state is $i$. Initially, $f[0] = \text{true}$, and the answer is $f[2^n - 1]$, where $n$ is the length of the array $\textit{nums}$. Additionally, we define $cur[i]$ to represent the sum of the last subset when the current selected numbers' state is $i$.
+
+We enumerate the states $i$ in the range $[0, 2^n]$. For each state $i$, if $f[i]$ is $\text{false}$, we skip it. Otherwise, we enumerate any number $\textit{nums}[j]$ in the array $\textit{nums}$. If $cur[i] + \textit{nums}[j] > s$, we break the enumeration loop because the subsequent numbers are larger and cannot be placed in the current subset. Otherwise, if the $j$-th bit of the binary representation of $i$ is $0$, it means the current $\textit{nums}[j]$ has not been selected. We can place it in the current subset, change the state to $i | 2^j$, update $cur[i | 2^j] = (cur[i] + \textit{nums}[j]) \bmod s$, and set $f[i | 2^j] = \text{true}$.
+
+Finally, we return $f[2^n - 1]$.
+
+The time complexity is $O(n \times 2^n)$, and the space complexity is $O(2^n)$. Here, $n$ represents the length of the array $\textit{nums}$.
 
 <!-- tabs:start -->
 
@@ -549,6 +624,38 @@ func canPartitionKSubsets(nums []int, k int) bool {
 }
 ```
 
+#### TypeScript
+
+```ts
+function canPartitionKSubsets(nums: number[], k: number): boolean {
+    let s = nums.reduce((a, b) => a + b);
+    if (s % k !== 0) {
+        return false;
+    }
+    s /= k;
+    nums.sort((a, b) => a - b);
+    const n = nums.length;
+    const f: boolean[] = Array(1 << n).fill(false);
+    f[0] = true;
+    const cur: number[] = Array(n).fill(0);
+    for (let i = 0; i < 1 << n; ++i) {
+        if (!f[i]) {
+            continue;
+        }
+        for (let j = 0; j < n; ++j) {
+            if (cur[i] + nums[j] > s) {
+                break;
+            }
+            if (((i >> j) & 1) === 0) {
+                f[i | (1 << j)] = true;
+                cur[i | (1 << j)] = (cur[i] + nums[j]) % s;
+            }
+        }
+    }
+    return f[(1 << n) - 1];
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->
diff --git a/solution/0600-0699/0698.Partition to K Equal Sum Subsets/Solution.cpp b/solution/0600-0699/0698.Partition to K Equal Sum Subsets/Solution.cpp
index 4624b6ac3e12c..07140a6d0f88c 100644
--- a/solution/0600-0699/0698.Partition to K Equal Sum Subsets/Solution.cpp	
+++ b/solution/0600-0699/0698.Partition to K Equal Sum Subsets/Solution.cpp	
@@ -8,8 +8,7 @@ class Solution {
         s /= k;
         int n = nums.size();
         vector<int> cur(k);
-        function<bool(int)> dfs;
-        dfs = [&](int i) {
+        function<bool(int)> dfs = [&](int i) {
             if (i == n) {
                 return true;
             }
@@ -28,4 +27,4 @@ class Solution {
         sort(nums.begin(), nums.end(), greater<int>());
         return dfs(0);
     }
-};
\ No newline at end of file
+};
diff --git a/solution/0600-0699/0698.Partition to K Equal Sum Subsets/Solution.py b/solution/0600-0699/0698.Partition to K Equal Sum Subsets/Solution.py
index 28bb91b8043da..a408312d49ee9 100644
--- a/solution/0600-0699/0698.Partition to K Equal Sum Subsets/Solution.py	
+++ b/solution/0600-0699/0698.Partition to K Equal Sum Subsets/Solution.py	
@@ -1,6 +1,6 @@
 class Solution:
     def canPartitionKSubsets(self, nums: List[int], k: int) -> bool:
-        def dfs(i):
+        def dfs(i: int) -> bool:
             if i == len(nums):
                 return True
             for j in range(k):
diff --git a/solution/0600-0699/0698.Partition to K Equal Sum Subsets/Solution.ts b/solution/0600-0699/0698.Partition to K Equal Sum Subsets/Solution.ts
index 69d688c6dbc92..236fc2b85c0e4 100644
--- a/solution/0600-0699/0698.Partition to K Equal Sum Subsets/Solution.ts	
+++ b/solution/0600-0699/0698.Partition to K Equal Sum Subsets/Solution.ts	
@@ -1,27 +1,28 @@
 function canPartitionKSubsets(nums: number[], k: number): boolean {
-    let s = nums.reduce((a, b) => a + b);
-    if (s % k !== 0) {
-        return false;
-    }
-    s /= k;
-    nums.sort((a, b) => a - b);
-    const n = nums.length;
-    const f: boolean[] = new Array(1 << n).fill(false);
-    f[0] = true;
-    const cur: number[] = new Array(n).fill(0);
-    for (let i = 0; i < 1 << n; ++i) {
-        if (!f[i]) {
-            continue;
+    const dfs = (i: number): boolean => {
+        if (i === nums.length) {
+            return true;
         }
-        for (let j = 0; j < n; ++j) {
-            if (cur[i] + nums[j] > s) {
-                break;
+        for (let j = 0; j < k; j++) {
+            if (j > 0 && cur[j] === cur[j - 1]) {
+                continue;
             }
-            if (((i >> j) & 1) === 0) {
-                f[i | (1 << j)] = true;
-                cur[i | (1 << j)] = (cur[i] + nums[j]) % s;
+            cur[j] += nums[i];
+            if (cur[j] <= s && dfs(i + 1)) {
+                return true;
             }
+            cur[j] -= nums[i];
         }
+        return false;
+    };
+
+    let s = nums.reduce((a, b) => a + b, 0);
+    const mod = s % k;
+    if (mod !== 0) {
+        return false;
     }
-    return f[(1 << n) - 1];
+    s = Math.floor(s / k);
+    const cur = Array(k).fill(0);
+    nums.sort((a, b) => b - a);
+    return dfs(0);
 }
diff --git a/solution/0600-0699/0698.Partition to K Equal Sum Subsets/Solution2.cpp b/solution/0600-0699/0698.Partition to K Equal Sum Subsets/Solution2.cpp
index 27aa6624efacc..706d16657cee2 100644
--- a/solution/0600-0699/0698.Partition to K Equal Sum Subsets/Solution2.cpp	
+++ b/solution/0600-0699/0698.Partition to K Equal Sum Subsets/Solution2.cpp	
@@ -10,8 +10,7 @@ class Solution {
         int n = nums.size();
         int mask = (1 << n) - 1;
         vector<int> f(1 << n);
-        function<bool(int, int)> dfs;
-        dfs = [&](int state, int t) {
+        function<bool(int, int)> dfs = [&](int state, int t) {
             if (state == mask) {
                 return true;
             }
@@ -35,4 +34,4 @@ class Solution {
         };
         return dfs(0, 0);
     }
-};
\ No newline at end of file
+};
diff --git a/solution/0600-0699/0698.Partition to K Equal Sum Subsets/Solution2.ts b/solution/0600-0699/0698.Partition to K Equal Sum Subsets/Solution2.ts
new file mode 100644
index 0000000000000..c9a183d059095
--- /dev/null
+++ b/solution/0600-0699/0698.Partition to K Equal Sum Subsets/Solution2.ts	
@@ -0,0 +1,36 @@
+function canPartitionKSubsets(nums: number[], k: number): boolean {
+    let s = nums.reduce((a, b) => a + b, 0);
+    if (s % k !== 0) {
+        return false;
+    }
+    s = Math.floor(s / k);
+    nums.sort((a, b) => a - b);
+    const n = nums.length;
+    const mask = (1 << n) - 1;
+    const f = Array(1 << n).fill(0);
+
+    const dfs = (state: number, t: number): boolean => {
+        if (state === mask) {
+            return true;
+        }
+        if (f[state] !== 0) {
+            return f[state] === 1;
+        }
+        for (let i = 0; i < n; ++i) {
+            if ((state >> i) & 1) {
+                continue;
+            }
+            if (t + nums[i] > s) {
+                break;
+            }
+            if (dfs(state | (1 << i), (t + nums[i]) % s)) {
+                f[state] = 1;
+                return true;
+            }
+        }
+        f[state] = -1;
+        return false;
+    };
+
+    return dfs(0, 0);
+}
diff --git a/solution/0600-0699/0698.Partition to K Equal Sum Subsets/Solution3.ts b/solution/0600-0699/0698.Partition to K Equal Sum Subsets/Solution3.ts
new file mode 100644
index 0000000000000..2d1d28df96bc8
--- /dev/null
+++ b/solution/0600-0699/0698.Partition to K Equal Sum Subsets/Solution3.ts	
@@ -0,0 +1,27 @@
+function canPartitionKSubsets(nums: number[], k: number): boolean {
+    let s = nums.reduce((a, b) => a + b);
+    if (s % k !== 0) {
+        return false;
+    }
+    s /= k;
+    nums.sort((a, b) => a - b);
+    const n = nums.length;
+    const f: boolean[] = Array(1 << n).fill(false);
+    f[0] = true;
+    const cur: number[] = Array(n).fill(0);
+    for (let i = 0; i < 1 << n; ++i) {
+        if (!f[i]) {
+            continue;
+        }
+        for (let j = 0; j < n; ++j) {
+            if (cur[i] + nums[j] > s) {
+                break;
+            }
+            if (((i >> j) & 1) === 0) {
+                f[i | (1 << j)] = true;
+                cur[i | (1 << j)] = (cur[i] + nums[j]) % s;
+            }
+        }
+    }
+    return f[(1 << n) - 1];
+}