diff --git a/solution/2900-2999/2967.Minimum Cost to Make Array Equalindromic/README.md b/solution/2900-2999/2967.Minimum Cost to Make Array Equalindromic/README.md
index d72a88ca8090e..2c4c5cf39502c 100644
--- a/solution/2900-2999/2967.Minimum Cost to Make Array Equalindromic/README.md	
+++ b/solution/2900-2999/2967.Minimum Cost to Make Array Equalindromic/README.md	
@@ -64,6 +64,14 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：预处理 + 排序 + 二分查找**
+
+题目中回文数的范围是 $[1, 10^9]$，回文数由于对称性，我们可以在 $[1, 10^5]$ 的范围内枚举，然后将其翻转后拼接，得到所有的回文数，注意，如果是奇数长度的回文数，我们在翻转前要去掉最后一位。预处理得到的回文数数组记为 $ps$。我们对数组 $ps$ 进行排序。
+
+接下来，我们对数组 $nums$ 进行排序，然后取 $nums$ 的中位数 $x$，我们只需要通过二分查找，在回文数组 $ps$ 中，找到一个与 $x$ 最接近的数，然后计算 $nums$ 变成这个数的代价，即可得到答案。
+
+时间复杂度 $O(n \times \log n)$，空间复杂度 $O(M)$。其中 $n$ 是数组 $nums$ 的长度，而 $M$ 是回文数组 $ps$ 的长度。
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -71,7 +79,24 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+ps = []
+for i in range(1, 10**5 + 1):
+    s = str(i)
+    t1 = s[::-1]
+    t2 = s[:-1][::-1]
+    ps.append(int(s + t1))
+    ps.append(int(s + t2))
+ps.sort()
+
+
+class Solution:
+    def minimumCost(self, nums: List[int]) -> int:
+        def f(x: int) -> int:
+            return sum(abs(v - x) for v in nums)
+
+        nums.sort()
+        i = bisect_left(ps, nums[len(nums) // 2])
+        return min(f(ps[j]) for j in range(i - 1, i + 2) if 0 <= j < len(ps))
 ```
 
 ### **Java**
@@ -79,19 +104,190 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+public class Solution {
+    private static long[] ps;
+    private int[] nums;
+
+    static {
+        ps = new long[2 * (int) 1e5];
+        for (int i = 1; i <= 1e5; i++) {
+            String s = Integer.toString(i);
+            String t1 = new StringBuilder(s).reverse().toString();
+            String t2 = new StringBuilder(s.substring(0, s.length() - 1)).reverse().toString();
+            ps[2 * i - 2] = Long.parseLong(s + t1);
+            ps[2 * i - 1] = Long.parseLong(s + t2);
+        }
+        Arrays.sort(ps);
+    }
+
+    public long minimumCost(int[] nums) {
+        this.nums = nums;
+        Arrays.sort(nums);
+        int i = Arrays.binarySearch(ps, nums[nums.length / 2]);
+        i = i < 0 ? -i - 1 : i;
+        long ans = 1L << 60;
+        for (int j = i - 1; j <= i + 1; j++) {
+            if (0 <= j && j < ps.length) {
+                ans = Math.min(ans, f(ps[j]));
+            }
+        }
+        return ans;
+    }
+
+    private long f(long x) {
+        long ans = 0;
+        for (int v : nums) {
+            ans += Math.abs(v - x);
+        }
+        return ans;
+    }
+}
 ```
 
 ### **C++**
 
 ```cpp
-
+using ll = long long;
+
+ll ps[2 * 100000];
+
+int init = [] {
+    for (int i = 1; i <= 100000; i++) {
+        string s = to_string(i);
+        string t1 = s;
+        reverse(t1.begin(), t1.end());
+        string t2 = s.substr(0, s.length() - 1);
+        reverse(t2.begin(), t2.end());
+        ps[2 * i - 2] = stoll(s + t1);
+        ps[2 * i - 1] = stoll(s + t2);
+    }
+    sort(ps, ps + 2 * 100000);
+    return 0;
+}();
+
+class Solution {
+public:
+    long long minimumCost(vector<int>& nums) {
+        sort(nums.begin(), nums.end());
+        int i = lower_bound(ps, ps + 2 * 100000, nums[nums.size() / 2]) - ps;
+        auto f = [&](ll x) {
+            ll ans = 0;
+            for (int& v : nums) {
+                ans += abs(v - x);
+            }
+            return ans;
+        };
+        ll ans = LLONG_MAX;
+        for (int j = i - 1; j <= i + 1; j++) {
+            if (0 <= j && j < 2 * 100000) {
+                ans = min(ans, f(ps[j]));
+            }
+        }
+        return ans;
+    }
+};
 ```
 
 ### **Go**
 
 ```go
+var ps [2 * 100000]int64
+
+func init() {
+	for i := 1; i <= 100000; i++ {
+		s := strconv.Itoa(i)
+		t1 := reverseString(s)
+		t2 := reverseString(s[:len(s)-1])
+		ps[2*i-2], _ = strconv.ParseInt(s+t1, 10, 64)
+		ps[2*i-1], _ = strconv.ParseInt(s+t2, 10, 64)
+	}
+	sort.Slice(ps[:], func(i, j int) bool {
+		return ps[i] < ps[j]
+	})
+}
+
+func reverseString(s string) string {
+	cs := []rune(s)
+	for i, j := 0, len(cs)-1; i < j; i, j = i+1, j-1 {
+		cs[i], cs[j] = cs[j], cs[i]
+	}
+	return string(cs)
+}
+
+func minimumCost(nums []int) int64 {
+	sort.Ints(nums)
+	i := sort.Search(len(ps), func(i int) bool {
+		return ps[i] >= int64(nums[len(nums)/2])
+	})
+
+	f := func(x int64) int64 {
+		var ans int64
+		for _, v := range nums {
+			ans += int64(abs(int(x - int64(v))))
+		}
+		return ans
+	}
+
+	ans := int64(math.MaxInt64)
+	for j := i - 1; j <= i+1; j++ {
+		if 0 <= j && j < len(ps) {
+			ans = min(ans, f(ps[j]))
+		}
+	}
+	return ans
+}
+
+func abs(x int) int {
+	if x < 0 {
+		return -x
+	}
+	return x
+}
+```
 
+### **TypeScript**
+
+```ts
+const ps = Array(2e5).fill(0);
+
+const init = (() => {
+    for (let i = 1; i <= 1e5; ++i) {
+        const s: string = i.toString();
+        const t1: string = s.split('').reverse().join('');
+        const t2: string = s.slice(0, -1).split('').reverse().join('');
+        ps[2 * i - 2] = parseInt(s + t1, 10);
+        ps[2 * i - 1] = parseInt(s + t2, 10);
+    }
+    ps.sort((a, b) => a - b);
+})();
+
+function minimumCost(nums: number[]): number {
+    const search = (x: number): number => {
+        let [l, r] = [0, ps.length];
+        while (l < r) {
+            const mid = (l + r) >> 1;
+            if (ps[mid] >= x) {
+                r = mid;
+            } else {
+                l = mid + 1;
+            }
+        }
+        return l;
+    };
+    const f = (x: number): number => {
+        return nums.reduce((acc, v) => acc + Math.abs(v - x), 0);
+    };
+
+    nums.sort((a, b) => a - b);
+    const i: number = search(nums[nums.length >> 1]);
+    let ans: number = Number.MAX_SAFE_INTEGER;
+    for (let j = i - 1; j <= i + 1; j++) {
+        if (j >= 0 && j < ps.length) {
+            ans = Math.min(ans, f(ps[j]));
+        }
+    }
+    return ans;
+}
 ```
 
 ### **...**
diff --git a/solution/2900-2999/2967.Minimum Cost to Make Array Equalindromic/README_EN.md b/solution/2900-2999/2967.Minimum Cost to Make Array Equalindromic/README_EN.md
index 2e9b90c469847..7d70b0530e373 100644
--- a/solution/2900-2999/2967.Minimum Cost to Make Array Equalindromic/README_EN.md	
+++ b/solution/2900-2999/2967.Minimum Cost to Make Array Equalindromic/README_EN.md	
@@ -58,30 +58,226 @@ It can be shown that changing all elements to any palindromic number other than
 
 ## Solutions
 
+**Solution 1: Preprocessing + Sorting + Binary Search**
+
+The range of palindrome numbers in the problem is $[1, 10^9]$. Due to the symmetry of palindrome numbers, we can enumerate in the range of $[1, 10^5]$, then reverse and concatenate them to get all palindrome numbers. Note that if it is an odd-length palindrome number, we need to remove the last digit before reversing. The array of palindrome numbers obtained by preprocessing is denoted as $ps$. We sort the array $ps$.
+
+Next, we sort the array $nums$ and take the median $x$ of $nums$. We only need to find a number in the palindrome array $ps$ that is closest to $x$ through binary search, and then calculate the cost of $nums$ becoming this number to get the answer.
+
+The time complexity is $O(n \times \log n)$, and the space complexity is $O(M)$. Here, $n$ is the length of the array $nums$, and $M$ is the length of the palindrome array $ps$.
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 ```python
-
+ps = []
+for i in range(1, 10**5 + 1):
+    s = str(i)
+    t1 = s[::-1]
+    t2 = s[:-1][::-1]
+    ps.append(int(s + t1))
+    ps.append(int(s + t2))
+ps.sort()
+
+
+class Solution:
+    def minimumCost(self, nums: List[int]) -> int:
+        def f(x: int) -> int:
+            return sum(abs(v - x) for v in nums)
+
+        nums.sort()
+        i = bisect_left(ps, nums[len(nums) // 2])
+        return min(f(ps[j]) for j in range(i - 1, i + 2) if 0 <= j < len(ps))
 ```
 
 ### **Java**
 
 ```java
-
+public class Solution {
+    private static long[] ps;
+    private int[] nums;
+
+    static {
+        ps = new long[2 * (int) 1e5];
+        for (int i = 1; i <= 1e5; i++) {
+            String s = Integer.toString(i);
+            String t1 = new StringBuilder(s).reverse().toString();
+            String t2 = new StringBuilder(s.substring(0, s.length() - 1)).reverse().toString();
+            ps[2 * i - 2] = Long.parseLong(s + t1);
+            ps[2 * i - 1] = Long.parseLong(s + t2);
+        }
+        Arrays.sort(ps);
+    }
+
+    public long minimumCost(int[] nums) {
+        this.nums = nums;
+        Arrays.sort(nums);
+        int i = Arrays.binarySearch(ps, nums[nums.length / 2]);
+        i = i < 0 ? -i - 1 : i;
+        long ans = 1L << 60;
+        for (int j = i - 1; j <= i + 1; j++) {
+            if (0 <= j && j < ps.length) {
+                ans = Math.min(ans, f(ps[j]));
+            }
+        }
+        return ans;
+    }
+
+    private long f(long x) {
+        long ans = 0;
+        for (int v : nums) {
+            ans += Math.abs(v - x);
+        }
+        return ans;
+    }
+}
 ```
 
 ### **C++**
 
 ```cpp
-
+using ll = long long;
+
+ll ps[2 * 100000];
+
+int init = [] {
+    for (int i = 1; i <= 100000; i++) {
+        string s = to_string(i);
+        string t1 = s;
+        reverse(t1.begin(), t1.end());
+        string t2 = s.substr(0, s.length() - 1);
+        reverse(t2.begin(), t2.end());
+        ps[2 * i - 2] = stoll(s + t1);
+        ps[2 * i - 1] = stoll(s + t2);
+    }
+    sort(ps, ps + 2 * 100000);
+    return 0;
+}();
+
+class Solution {
+public:
+    long long minimumCost(vector<int>& nums) {
+        sort(nums.begin(), nums.end());
+        int i = lower_bound(ps, ps + 2 * 100000, nums[nums.size() / 2]) - ps;
+        auto f = [&](ll x) {
+            ll ans = 0;
+            for (int& v : nums) {
+                ans += abs(v - x);
+            }
+            return ans;
+        };
+        ll ans = LLONG_MAX;
+        for (int j = i - 1; j <= i + 1; j++) {
+            if (0 <= j && j < 2 * 100000) {
+                ans = min(ans, f(ps[j]));
+            }
+        }
+        return ans;
+    }
+};
 ```
 
 ### **Go**
 
 ```go
+var ps [2 * 100000]int64
+
+func init() {
+	for i := 1; i <= 100000; i++ {
+		s := strconv.Itoa(i)
+		t1 := reverseString(s)
+		t2 := reverseString(s[:len(s)-1])
+		ps[2*i-2], _ = strconv.ParseInt(s+t1, 10, 64)
+		ps[2*i-1], _ = strconv.ParseInt(s+t2, 10, 64)
+	}
+	sort.Slice(ps[:], func(i, j int) bool {
+		return ps[i] < ps[j]
+	})
+}
+
+func reverseString(s string) string {
+	cs := []rune(s)
+	for i, j := 0, len(cs)-1; i < j; i, j = i+1, j-1 {
+		cs[i], cs[j] = cs[j], cs[i]
+	}
+	return string(cs)
+}
+
+func minimumCost(nums []int) int64 {
+	sort.Ints(nums)
+	i := sort.Search(len(ps), func(i int) bool {
+		return ps[i] >= int64(nums[len(nums)/2])
+	})
+
+	f := func(x int64) int64 {
+		var ans int64
+		for _, v := range nums {
+			ans += int64(abs(int(x - int64(v))))
+		}
+		return ans
+	}
+
+	ans := int64(math.MaxInt64)
+	for j := i - 1; j <= i+1; j++ {
+		if 0 <= j && j < len(ps) {
+			ans = min(ans, f(ps[j]))
+		}
+	}
+	return ans
+}
+
+func abs(x int) int {
+	if x < 0 {
+		return -x
+	}
+	return x
+}
+```
 
+### **TypeScript**
+
+```ts
+const ps = Array(2e5).fill(0);
+
+const init = (() => {
+    for (let i = 1; i <= 1e5; ++i) {
+        const s: string = i.toString();
+        const t1: string = s.split('').reverse().join('');
+        const t2: string = s.slice(0, -1).split('').reverse().join('');
+        ps[2 * i - 2] = parseInt(s + t1, 10);
+        ps[2 * i - 1] = parseInt(s + t2, 10);
+    }
+    ps.sort((a, b) => a - b);
+})();
+
+function minimumCost(nums: number[]): number {
+    const search = (x: number): number => {
+        let [l, r] = [0, ps.length];
+        while (l < r) {
+            const mid = (l + r) >> 1;
+            if (ps[mid] >= x) {
+                r = mid;
+            } else {
+                l = mid + 1;
+            }
+        }
+        return l;
+    };
+    const f = (x: number): number => {
+        return nums.reduce((acc, v) => acc + Math.abs(v - x), 0);
+    };
+
+    nums.sort((a, b) => a - b);
+    const i: number = search(nums[nums.length >> 1]);
+    let ans: number = Number.MAX_SAFE_INTEGER;
+    for (let j = i - 1; j <= i + 1; j++) {
+        if (j >= 0 && j < ps.length) {
+            ans = Math.min(ans, f(ps[j]));
+        }
+    }
+    return ans;
+}
 ```
 
 ### **...**
diff --git a/solution/2900-2999/2967.Minimum Cost to Make Array Equalindromic/Solution.cpp b/solution/2900-2999/2967.Minimum Cost to Make Array Equalindromic/Solution.cpp
new file mode 100644
index 0000000000000..923c31f196570
--- /dev/null
+++ b/solution/2900-2999/2967.Minimum Cost to Make Array Equalindromic/Solution.cpp	
@@ -0,0 +1,39 @@
+using ll = long long;
+
+ll ps[2 * 100000];
+
+int init = [] {
+    for (int i = 1; i <= 100000; i++) {
+        string s = to_string(i);
+        string t1 = s;
+        reverse(t1.begin(), t1.end());
+        string t2 = s.substr(0, s.length() - 1);
+        reverse(t2.begin(), t2.end());
+        ps[2 * i - 2] = stoll(s + t1);
+        ps[2 * i - 1] = stoll(s + t2);
+    }
+    sort(ps, ps + 2 * 100000);
+    return 0;
+}();
+
+class Solution {
+public:
+    long long minimumCost(vector<int>& nums) {
+        sort(nums.begin(), nums.end());
+        int i = lower_bound(ps, ps + 2 * 100000, nums[nums.size() / 2]) - ps;
+        auto f = [&](ll x) {
+            ll ans = 0;
+            for (int& v : nums) {
+                ans += abs(v - x);
+            }
+            return ans;
+        };
+        ll ans = LLONG_MAX;
+        for (int j = i - 1; j <= i + 1; j++) {
+            if (0 <= j && j < 2 * 100000) {
+                ans = min(ans, f(ps[j]));
+            }
+        }
+        return ans;
+    }
+};
\ No newline at end of file
diff --git a/solution/2900-2999/2967.Minimum Cost to Make Array Equalindromic/Solution.go b/solution/2900-2999/2967.Minimum Cost to Make Array Equalindromic/Solution.go
new file mode 100644
index 0000000000000..3f6b36621dcbd
--- /dev/null
+++ b/solution/2900-2999/2967.Minimum Cost to Make Array Equalindromic/Solution.go	
@@ -0,0 +1,52 @@
+var ps [2 * 100000]int64
+
+func init() {
+	for i := 1; i <= 100000; i++ {
+		s := strconv.Itoa(i)
+		t1 := reverseString(s)
+		t2 := reverseString(s[:len(s)-1])
+		ps[2*i-2], _ = strconv.ParseInt(s+t1, 10, 64)
+		ps[2*i-1], _ = strconv.ParseInt(s+t2, 10, 64)
+	}
+	sort.Slice(ps[:], func(i, j int) bool {
+		return ps[i] < ps[j]
+	})
+}
+
+func reverseString(s string) string {
+	cs := []rune(s)
+	for i, j := 0, len(cs)-1; i < j; i, j = i+1, j-1 {
+		cs[i], cs[j] = cs[j], cs[i]
+	}
+	return string(cs)
+}
+
+func minimumCost(nums []int) int64 {
+	sort.Ints(nums)
+	i := sort.Search(len(ps), func(i int) bool {
+		return ps[i] >= int64(nums[len(nums)/2])
+	})
+
+	f := func(x int64) int64 {
+		var ans int64
+		for _, v := range nums {
+			ans += int64(abs(int(x - int64(v))))
+		}
+		return ans
+	}
+
+	ans := int64(math.MaxInt64)
+	for j := i - 1; j <= i+1; j++ {
+		if 0 <= j && j < len(ps) {
+			ans = min(ans, f(ps[j]))
+		}
+	}
+	return ans
+}
+
+func abs(x int) int {
+	if x < 0 {
+		return -x
+	}
+	return x
+}
\ No newline at end of file
diff --git a/solution/2900-2999/2967.Minimum Cost to Make Array Equalindromic/Solution.java b/solution/2900-2999/2967.Minimum Cost to Make Array Equalindromic/Solution.java
new file mode 100644
index 0000000000000..a5fcc83957b9c
--- /dev/null
+++ b/solution/2900-2999/2967.Minimum Cost to Make Array Equalindromic/Solution.java	
@@ -0,0 +1,38 @@
+public class Solution {
+    private static long[] ps;
+    private int[] nums;
+
+    static {
+        ps = new long[2 * (int) 1e5];
+        for (int i = 1; i <= 1e5; i++) {
+            String s = Integer.toString(i);
+            String t1 = new StringBuilder(s).reverse().toString();
+            String t2 = new StringBuilder(s.substring(0, s.length() - 1)).reverse().toString();
+            ps[2 * i - 2] = Long.parseLong(s + t1);
+            ps[2 * i - 1] = Long.parseLong(s + t2);
+        }
+        Arrays.sort(ps);
+    }
+
+    public long minimumCost(int[] nums) {
+        this.nums = nums;
+        Arrays.sort(nums);
+        int i = Arrays.binarySearch(ps, nums[nums.length / 2]);
+        i = i < 0 ? -i - 1 : i;
+        long ans = 1L << 60;
+        for (int j = i - 1; j <= i + 1; j++) {
+            if (0 <= j && j < ps.length) {
+                ans = Math.min(ans, f(ps[j]));
+            }
+        }
+        return ans;
+    }
+
+    private long f(long x) {
+        long ans = 0;
+        for (int v : nums) {
+            ans += Math.abs(v - x);
+        }
+        return ans;
+    }
+}
\ No newline at end of file
diff --git a/solution/2900-2999/2967.Minimum Cost to Make Array Equalindromic/Solution.py b/solution/2900-2999/2967.Minimum Cost to Make Array Equalindromic/Solution.py
new file mode 100644
index 0000000000000..602f8df4810ee
--- /dev/null
+++ b/solution/2900-2999/2967.Minimum Cost to Make Array Equalindromic/Solution.py	
@@ -0,0 +1,18 @@
+ps = []
+for i in range(1, 10**5 + 1):
+    s = str(i)
+    t1 = s[::-1]
+    t2 = s[:-1][::-1]
+    ps.append(int(s + t1))
+    ps.append(int(s + t2))
+ps.sort()
+
+
+class Solution:
+    def minimumCost(self, nums: List[int]) -> int:
+        def f(x: int) -> int:
+            return sum(abs(v - x) for v in nums)
+
+        nums.sort()
+        i = bisect_left(ps, nums[len(nums) // 2])
+        return min(f(ps[j]) for j in range(i - 1, i + 2) if 0 <= j < len(ps))
diff --git a/solution/2900-2999/2967.Minimum Cost to Make Array Equalindromic/Solution.ts b/solution/2900-2999/2967.Minimum Cost to Make Array Equalindromic/Solution.ts
new file mode 100644
index 0000000000000..431112126f4db
--- /dev/null
+++ b/solution/2900-2999/2967.Minimum Cost to Make Array Equalindromic/Solution.ts	
@@ -0,0 +1,40 @@
+const ps = Array(2e5).fill(0);
+
+const init = (() => {
+    for (let i = 1; i <= 1e5; ++i) {
+        const s: string = i.toString();
+        const t1: string = s.split('').reverse().join('');
+        const t2: string = s.slice(0, -1).split('').reverse().join('');
+        ps[2 * i - 2] = parseInt(s + t1, 10);
+        ps[2 * i - 1] = parseInt(s + t2, 10);
+    }
+    ps.sort((a, b) => a - b);
+})();
+
+function minimumCost(nums: number[]): number {
+    const search = (x: number): number => {
+        let [l, r] = [0, ps.length];
+        while (l < r) {
+            const mid = (l + r) >> 1;
+            if (ps[mid] >= x) {
+                r = mid;
+            } else {
+                l = mid + 1;
+            }
+        }
+        return l;
+    };
+    const f = (x: number): number => {
+        return nums.reduce((acc, v) => acc + Math.abs(v - x), 0);
+    };
+
+    nums.sort((a, b) => a - b);
+    const i: number = search(nums[nums.length >> 1]);
+    let ans: number = Number.MAX_SAFE_INTEGER;
+    for (let j = i - 1; j <= i + 1; j++) {
+        if (j >= 0 && j < ps.length) {
+            ans = Math.min(ans, f(ps[j]));
+        }
+    }
+    return ans;
+}