Minimum indexed character solutoin added

Author: Vipullakum007

dsa-solutions/gfg-solutions/0024-minimum-indexed-character.md

@@ -0,0 +1,180 @@
+---
+id: minimum-indexed-character
+title: Minimum Indexed Character
+sidebar_label: 0024 Minimum Indexed Character
+tags:
+- String
+- Hashing
+- JavaScript
+- TypeScript
+- Python
+- Java
+- C++
+description: "This document explores different approaches to solving the problem of finding the minimum index of a character in a string that is also present in another string, including solutions in JavaScript, TypeScript, Python, Java, and C++."
+---
+
+## Problem
+
+Given a string `str` and another string `patt`, find the minimum index of the character in `str` that is also present in `patt`.
+
+### Examples
+
+**Example 1:**
+
+```
+Input:
+str = "geeksforgeeks"
+patt = "set"
+
+Output:
+1
+
+Explanation:
+'e' is the character which is present in the given string "geeksforgeeks" and is also present in "set". The minimum index of 'e' is 1.
+```
+
+**Example 2:**
+
+```
+Input:
+str = "adcffaet"
+patt = "onkl"
+
+Output:
+-1
+
+Explanation:
+There are no characters that are common in "patt" and "str".
+```
+
+### Your Task
+You only need to complete the function `minIndexChar()` that returns the index of the answer in `str` or returns `-1` if no character of `patt` is present in `str`.
+
+**Expected Time Complexity:** O(N)  
+**Expected Auxiliary Space:** O(Number of distinct characters)
+
+### Constraints
+- 1 ≤ |str|,|patt| ≤ 10^5
+- 'a' ≤ str[i], patt[i] ≤ 'z'
+
+## Solution
+
+### Approach
+
+To solve this problem, we can follow these steps:
+
+1. Iterate through the characters in the `patt` string.
+2. For each character in `patt`, find its index in the `str` string using the `find` method.
+3. Store the indices in a vector.
+4. Iterate through the vector to find the minimum index that is not `-1` (indicating the character is found in `str`).
+5. If no valid index is found, return `-1`; otherwise, return the minimum index.
+
+### Implementation
+
+<Tabs>
+  <TabItem value="cpp" label="C++">
+
+```cpp
+class Solution
+{
+  public:
+    int minIndexChar(string str, string patt)
+    {
+        int res = INT_MAX;
+        vector<int> v1;
+        for (int i = 0; i < patt.length(); i++)
+        {
+            int p = str.find(patt[i]);
+            v1.push_back(p);
+        }
+        int min = INT_MAX;
+        for (int i = 0; i < v1.size(); i++)
+        {
+            if (min > v1[i] && v1[i] != -1)
+                min = v1[i];
+        }
+        if (min == INT_MAX)
+            return -1;
+        return min;
+    }
+};
+```
+
+  </TabItem>
+  <TabItem value="javascript" label="JavaScript">
+
+```javascript
+function minIndexChar(str, patt) {
+    let minIndex = Infinity;
+    for (let char of patt) {
+        let index = str.indexOf(char);
+        if (index !== -1 && index < minIndex) {
+            minIndex = index;
+        }
+    }
+    return minIndex === Infinity ? -1 : minIndex;
+}
+```
+
+  </TabItem>
+  <TabItem value="typescript" label="TypeScript">
+
+```typescript
+function minIndexChar(str: string, patt: string): number {
+    let minIndex = Infinity;
+    for (let char of patt) {
+        let index = str.indexOf(char);
+        if (index !== -1 && index < minIndex) {
+            minIndex = index;
+        }
+    }
+    return minIndex === Infinity ? -1 : minIndex;
+}
+```
+
+  </TabItem>
+  <TabItem value="python" label="Python">
+
+```python
+class Solution:
+    def minIndexChar(self, str: str, patt: str) -> int:
+        min_index = float('inf')
+        for char in patt:
+            index = str.find(char)
+            if index != -1 and index < min_index:
+                min_index = index
+        return -1 if min_index == float('inf') else min_index
+```
+
+  </TabItem>
+  <TabItem value="java" label="Java">
+
+```java
+class Solution {
+    public int minIndexChar(String str, String patt) {
+        int minIndex = Integer.MAX_VALUE;
+        for (char ch : patt.toCharArray()) {
+            int index = str.indexOf(ch);
+            if (index != -1 && index < minIndex) {
+                minIndex = index;
+            }
+        }
+        return minIndex == Integer.MAX_VALUE ? -1 : minIndex;
+    }
+}
+```
+
+  </TabItem>
+</Tabs>
+
+### Complexity Analysis
+
+- **Time Complexity:** O(N), where N is the length of the string `str`. We iterate through each character in `patt` and use the `find` or `indexOf` method, which runs in O(N) time.
+- **Space Complexity:** O(1), as we only use a constant amount of extra space for variables.
+
+---
+
+## References
+
+- **GeeksforGeeks Problem:** [Minimum Indexed Character](https://www.geeksforgeeks.org/minimum-indexed-character/)
+- **Authors GeeksforGeeks Profile:** [Vipul lakum](https://www.geeksforgeeks.org/user/lakumvipwjge/)