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# Keep track of previous indexes after sorting a vector in C++ STL

• Difficulty Level : Medium
• Last Updated : 06 Mar, 2019

Prerequisite: vector, vector pair sorting

Given a vector, keep track of the present indexes corresponding to each element and after sorting print element with its previous respective indexes.

Examples:

Input: Arr[] = {2, 5, 3, 7, 1}
Output: {1, 4} {2, 0} {3, 2} {5, 1} {7, 3}
Explanation:
Before sorting [index(element)]: [0(2), 1(5), 2(3), 3(7), 4(1)]
After sorting [previous_index(element)]: [4(1), 0(2), 2(3), 1(5), 3(7)]

Input: Arr[] = {4, 5, 10, 8, 3, 11}
Output: {3, 4} {4, 0} {5, 1} {8, 3} {10, 2} {11, 5}

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea is to store each element with its present index in a vector pair and then sort all the elements of the vector, Finally, print the elements with its index associated with it.

Below is the implementation of the above approach:

 // C++ implementation to keep track// of previous indexes// after sorting a vector  #include using namespace std;  void sortArr(int arr[], int n){      // Vector to store element    // with respective present index    vector > vp;      // Inserting element in pair vector    // to keep track of previous indexes    for (int i = 0; i < n; ++i) {        vp.push_back(make_pair(arr[i], i));    }      // Sorting pair vector    sort(vp.begin(), vp.end());      // Displaying sorted element    // with previous indexes    // corresponding to each element    cout << "Element\t"         << "index" << endl;    for (int i = 0; i < vp.size(); i++) {        cout << vp[i].first << "\t"             << vp[i].second << endl;    }}  // Driver codeint main(){    int arr[] = { 2, 5, 3, 7, 1 };    int n = sizeof(arr) / sizeof(arr[0]);    sortArr(arr, n);      return 0;}
Output:
Element    index
1    4
2    0
3    2
5    1
7    3

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