value equal to index value

Author: iamanolive

dsa-solutions/gfg-solutions/0001-value-equal-to-index-value.md

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+---
+id: value-equal-to-index-value
+title: Value Equal to Index Value Problem (Geeks for Geeks)
+sidebar_label: 0001 - Value Equal to Index Value 
+tags:
+  - Beginner
+  - Index Values
+  - Array
+  - Geeks for Geeks
+  - CPP
+  - Python
+  - DSA
+description: "This is a solution to the Value Equal to Index Value problem on Geeks for Geeks."
+---
+
+## Problem Description
+
+Given an array Arr of N positive integers. Your task is to find the elements whose value is equal to that of its index value ( Consider 1-based indexing ).<br />
+
+Note: There can be more than one element in the array which have the same value as its index. You need to include every such element's index. Follows 1-based indexing of the array.
+
+## Examples
+
+Example 1:
+
+```
+Input:
+N = 5
+Arr[] = {15, 2, 45, 12, 7}
+Output: 2
+Explanation: Only Arr[2] = 2 exists here.
+```
+
+Example 2:
+
+```
+Input: 
+N = 1
+Arr[] = {1}
+Output: 1
+Explanation: Here Arr[1] = 1 exists.
+```
+
+## Your Task
+
+You don't need to read input or print anything. Your task is to complete the function valueEqualToIndex() which takes the array of integers arr[] and n as parameters and returns an array of indices where the given conditions are satisfied. When there is no such element exists then return an empty array of length 0. For C users, you need to modify the array(arr), in place such that the indices not in the final answer should be marked 0.
+
+## Constraints
+
+* 1 ≤ N ≤ 105
+* 1 ≤ `Arr[i]` ≤ 106
+
+## Problem Explanation
+
+The problem involves finding elements in an array whose values match their index positions, considering 1-based indexing. This means if an element's value matches its position in the array when counting from 1 (not 0), it should be included in the result.<br />
+
+Example:
+
+* Given an array [15, 2, 45, 12, 7]:
+    - At 0-based index 0, the value is 15 (not equal to 1)
+    - At 0-based index 1, the value is 2 (equal to 2)
+    - At 0-based index 2, the value is 45 (not equal to 3)
+    - At 0-based index 3, the value is 12 (not equal to 4)
+    - At 0-based index 4, the value is 7 (not equal to 5)
+
+The only element that matches its index is 2 at index 2.
+
+## Python Solution Code
+
+```py
+class Solution:
+    
+    def valueEqualToIndex(self, arr, n):
+        result = []
+        for i in range(n):
+            if (arr[i] == i + 1):
+                result.append(arr[i]);
+        return result
+```
+
+## C++ Solution Code
+
+```cpp
+class Solution {
+    public:
+    	vector<int> valueEqualToIndex(int arr[], int n) {
+    	    vector<int> result;
+    	    for(int i = 0; i < n; i++) {
+    	        if(arr[i] == i + 1)
+    	            result.push_back(arr[i]);
+    	    } return result;
+        }
+};
+```
+
+## Solution Logic
+
+1. Initialize an empty result list: This will store the elements that satisfy the condition.
+2. Iterate through the array: Use a loop to go through each element in the array.
+3. Check the condition: For each element, check if the element's value matches its 1-based index. Since array indexing typically starts at 0 in most programming languages, you adjust by adding 1 to the index.
+4. Store matching elements: If an element's value matches its 1-based index, add it to the result list.
+5. Return the result list: After checking all elements, return the list of elements that met the condition.
+
+## Example Walkthrough
+
+For the array `[15, 2, 45, 12, 7]` with length `n = 5`:
+
+* Initialize an empty list: `result = []`
+* Loop through the array with indices from 0 to 4:
+    - Check element at index 0 (1-based index 1): 15 ≠ 1
+    Check element at index 1 (1-based index 2): 2 = 2 (add to result)
+    Check element at index 2 (1-based index 3): 45 ≠ 3
+    Check element at index 3 (1-based index 4): 12 ≠ 4
+    Check element at index 4 (1-based index 5): 7 ≠ 5
+
+Final `result` list: `[2]`
+
+## Time Complexity
+
+The time complexity of this solution is O(N) where N is the number of elements in the array. This is because the solution involves a single pass through the array to check each element against its 1-based index.
+
+## Space Complexity
+
+The space complexity is O(1) for the input space, and O(K) for the output space, where K is the number of elements matching the condition. In the worst case, if all elements match, the space used by the result list will be O(N).

dsa-solutions/gfg-solutions/_category_.json

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+{
+  "label": "LeetCode Solutions",
+  "position": 4,
+  "link": {
+    "type": "generated-index",
+    "description": "All Geeks for Geeks solutions in one place. Sorted by difficulty. Contains problem description, solution, and explanation."
+  }
+}