# Queries to find the minimum index in given array having at least value X

Given an array arr[] of size N and an array Q[] consisting of M integers, each representing a query, the task for each query Q[i] is to find the smallest index of an array element whose value is greater than or equal to Q[i]. If no such index exists, then print -1.

Examples:

Input: arr[] = { 1, 9 }, Q[] = { 7, 10, 0 }
Output: 1 -1 0
Explanation:
The smallest index of arr[] whose value is greater than Q[0] is 1.
No such index exists in arr[] whose value is greater than Q[1].
The smallest index of arr[] whose value is greater than Q[2] is 0.
Therefore, the required output is 1 -1 0.

Input: arr[] = {2, 3, 4}, Q[] = {2, 3, 4}
Output: 0 1 2

Approach:The problem can be solved using Binary search and Prefix Sum technique. Follow the steps below to solve the problem:

Below is the implementation of the above approach:

## C++

 `// C++ program to implement` `// the above approach` `#include ` `using` `namespace` `std;`   `// Function to find the smallest index` `// of an array element whose value is` `// less than or equal to Q[i]` `void` `minimumIndex(vector<``int``>& arr,` `                  ``vector<``int``>& Q)` `{`   `    ``// Stores size of array` `    ``int` `N = arr.size();`   `    ``// Stores count of queries` `    ``int` `M = Q.size();`   `    ``// Store array elements along` `    ``// with the index` `    ``vector > storeArrIdx;`   `    ``// Store smallest index of an array` `    ``// element whose value is greater` `    ``// than or equal to i` `    ``vector<``int``> minIdx(N);`   `    ``// Traverse the array` `    ``for` `(``int` `i = 0; i < N; ++i) {`   `        ``// Insert {arr[i], i} into` `        ``// storeArrIdx[]` `        ``storeArrIdx.push_back({ arr[i], i });` `    ``}`   `    ``// Sort the array` `    ``sort(arr.begin(), arr.end());`   `    ``// Sort the storeArrIdx` `    ``sort(storeArrIdx.begin(),` `         ``storeArrIdx.end());`   `    ``// Stores index of arr[N - 1] in` `    ``// sorted order` `    ``minIdx[N - 1]` `        ``= storeArrIdx[N - 1].second;`   `    ``// Traverse the array storeArrIdx[]` `    ``for` `(``int` `i = N - 2; i >= 0; i--) {`   `        ``// Update minIdx[i]` `        ``minIdx[i] = min(minIdx[i + 1],` `                        ``storeArrIdx[i].second);` `    ``}`   `    ``// Traverse the array Q[]` `    ``for` `(``int` `i = 0; i < M; i++) {`   `        ``// Store the index of` `        ``// lower_bound of Q[i]` `        ``int` `pos` `            ``= lower_bound(arr.begin(),` `                          ``arr.end(), Q[i])` `              ``- arr.begin();`   `        ``// If no index found whose value` `        ``// greater than or equal to arr[i]` `        ``if` `(pos == N) {` `            ``cout << -1 << ``" "``;` `            ``continue``;` `        ``}`   `        ``// Print smallest index whose value` `        ``// greater than or equal to Q[i]` `        ``cout << minIdx[pos] << ``" "``;` `    ``}` `}`   `// Driver Code` `int` `main()` `{`   `    ``vector<``int``> arr = { 1, 9 };` `    ``vector<``int``> Q = { 7, 10, 0 };`   `    ``minimumIndex(arr, Q);` `    ``return` `0;` `}`

## Java

 `// Java program for above approach` `import` `java.util.*;` `import` `java.lang.*;` `class` `pair` `{` `  ``int` `element,index;` `  ``pair(``int` `element, ``int` `index)` `  ``{` `    ``this``.element = element;` `    ``this``.index = index;` `  ``}` `}` `class` `GFG` `{`   `  ``// Function to find the smallest index` `  ``// of an array element whose value is` `  ``// less than or equal to Q[i]` `  ``static` `void` `minimumIndex(``int``[] arr,` `                           ``int``[] Q)` `  ``{`   `    ``// Stores size of array` `    ``int` `N = arr.length;`   `    ``// Stores count of queries` `    ``int` `M = Q.length;`   `    ``// Store array elements along` `    ``// with the index` `    ``ArrayList storeArrIdx = ``new` `ArrayList<>();`   `    ``// Store smallest index of an array` `    ``// element whose value is greater` `    ``// than or equal to i` `    ``int``[] minIdx = ``new` `int``[N];`   `    ``// Traverse the array` `    ``for` `(``int` `i = ``0``; i < N; ++i) ` `    ``{`   `      ``// Insert {arr[i], i} into` `      ``// storeArrIdx[]` `      ``storeArrIdx.add(``new` `pair(arr[i], i));` `    ``}`   `    ``// Sort the array` `    ``Arrays.sort(arr);`   `    ``// Sort the storeArrIdx` `    ``Collections.sort(storeArrIdx, (a, b)->a.element-b.element);`   `    ``// Stores index of arr[N - 1] in` `    ``// sorted order` `    ``minIdx[N - ``1``]` `      ``= storeArrIdx.get(N - ``1``).index;`   `    ``// Traverse the array storeArrIdx[]` `    ``for` `(``int` `i = N - ``2``; i >= ``0``; i--) {`   `      ``// Update minIdx[i]` `      ``minIdx[i] =Math.min(minIdx[i + ``1``],` `                          ``storeArrIdx.get(i).index);` `    ``}`   `    ``// Traverse the array Q[]` `    ``for` `(``int` `i = ``0``; i < M; i++) {`   `      ``// Store the index of` `      ``// lower_bound of Q[i]` `      ``int` `pos` `        ``= lower_bound(arr, Q[i]);`   `      ``// If no index found whose value` `      ``// greater than or equal to arr[i]` `      ``if` `(pos == N) {` `        ``System.out.print(``"-1"``+``" "``);` `        ``continue``;` `      ``}`   `      ``// Print smallest index whose value` `      ``// greater than or equal to Q[i]` `      ``System.out.print(minIdx[pos]+``" "``);` `    ``}` `  ``}` `  ``static` `int` `lower_bound(``int``[] arr,``int` `element)` `  ``{` `    ``for``(``int` `i = ``0``; i < arr.length; i++)` `      ``if``(element <= arr[i])` `        ``return` `i;`   `    ``return` `arr.length;  ` `  ``}`   `  ``// Driver function` `  ``public` `static` `void` `main (String[] args)` `  ``{` `    ``int``[] arr = { ``1``, ``9` `};` `    ``int``[] Q = { ``7``, ``10``, ``0` `};`   `    ``minimumIndex(arr, Q);` `  ``}` `}`   `// This code is contributed by offbeat`

## Python3

 `# Python3 program to implement` `# the above approachf` `from` `bisect ``import` `bisect_left`   `# Function to find the smallest index` `# of an array element whose value is` `# less than or equal to Q[i]` `def` `minimumIndex(arr, Q):`   `    ``# Stores size of array` `    ``N ``=` `len``(arr)`   `    ``# Stores count of queries` `    ``M ``=` `len``(Q)`   `    ``# Store array elements along` `    ``# with the index` `    ``storeArrIdx ``=` `[]`   `    ``# Store smallest index of an array` `    ``# element whose value is greater` `    ``# than or equal to i` `    ``minIdx ``=` `[``0``]``*``(N)`   `    ``# Traverse the array` `    ``for` `i ``in` `range``(N):` `      `  `        ``# Insert {arr[i], i} into` `        ``# storeArrIdx[]` `        ``storeArrIdx.append([arr[i], i])`   `    ``# Sort the array` `    ``arr ``=` `sorted``(arr)`   `    ``# Sort the storeArrIdx` `    ``storeArrIdx ``=` `sorted``(storeArrIdx)`   `    ``# Stores index of arr[N - 1] in` `    ``# sorted order` `    ``minIdx[N ``-` `1``] ``=` `storeArrIdx[N ``-` `1``][``1``]`   `    ``# Traverse the array storeArrIdx[]` `    ``for` `i ``in` `range``(N ``-` `2``, ``-``1``, ``-``1``):`   `        ``# Update minIdx[i]` `        ``minIdx[i] ``=` `min``(minIdx[i ``+` `1``], storeArrIdx[i][``1``])`   `    ``# Traverse the array Q[]` `    ``for` `i ``in` `range``(M):`   `        ``# Store the index of` `        ``# lower_bound of Q[i]` `        ``pos ``=` `bisect_left(arr, Q[i])`   `        ``# If no index found whose value` `        ``# greater than or equal to arr[i]` `        ``if` `(pos ``=``=` `N):` `            ``print``(``-``1``, end ``=` `" "``)` `            ``continue`   `        ``# Print smallest index whose value` `        ``# greater than or equal to Q[i]` `        ``print``(minIdx[pos], end ``=` `" "``)`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    ``arr ``=` `[``1``, ``9``]` `    ``Q ``=` `[``7``, ``10``, ``0``]` `    ``minimumIndex(arr, Q)`   `    ``# This code is contributed by mohit kumar 29`

## C#

 `// C# program for above approach`   `using` `System;` `using` `System.Collections.Generic;` `using` `System.Linq;`   `class` `pair` `{` `  ``public` `int` `element,index;` `  ``public` `pair(``int` `element, ``int` `index)` `  ``{` `    ``this``.element = element;` `    ``this``.index = index;` `  ``}` `}`   `class` `GFG` `{`   `  ``// Function to find the smallest index` `  ``// of an array element whose value is` `  ``// less than or equal to Q[i]` `  ``static` `void` `minimumIndex(``int``[] arr,` `                           ``int``[] Q)` `  ``{`   `    ``// Stores size of array` `    ``int` `N = arr.Length;`   `    ``// Stores count of queries` `    ``int` `M = Q.Length;`   `    ``// Store array elements along` `    ``// with the index` `    ``List storeArrIdx = ``new` `List();`   `    ``// Store smallest index of an array` `    ``// element whose value is greater` `    ``// than or equal to i` `    ``int``[] minIdx = ``new` `int``[N];`   `    ``// Traverse the array` `    ``for` `(``int` `i = 0; i < N; ++i) ` `    ``{`   `      ``// Insert {arr[i], i} into` `      ``// storeArrIdx[]` `      ``storeArrIdx.Add(``new` `pair(arr[i], i));` `    ``}`   `    ``// Sort the array` `    ``Array.Sort(arr);`   `    ``// Sort the storeArrIdx` `    ``storeArrIdx = storeArrIdx.OrderBy(a => a.element).ToList();`   `    ``// Stores index of arr[N - 1] in` `    ``// sorted order` `    ``minIdx[N - 1]` `      ``= storeArrIdx[N - 1].index;`   `    ``// Traverse the array storeArrIdx[]` `    ``for` `(``int` `i = N - 2; i >= 0; i--) {`   `      ``// Update minIdx[i]` `      ``minIdx[i] =Math.Min(minIdx[i + 1],` `                          ``storeArrIdx[i].index);` `    ``}`   `    ``// Traverse the array Q[]` `    ``for` `(``int` `i = 0; i < M; i++) {`   `      ``// Store the index of` `      ``// lower_bound of Q[i]` `      ``int` `pos` `        ``= lower_bound(arr, Q[i]);`   `      ``// If no index found whose value` `      ``// greater than or equal to arr[i]` `      ``if` `(pos == N) {` `        ``Console.Write(``"-1"``+``" "``);` `        ``continue``;` `      ``}`   `      ``// Print smallest index whose value` `      ``// greater than or equal to Q[i]` `      ``Console.Write(minIdx[pos]+``" "``);` `    ``}` `  ``}` `  ``static` `int` `lower_bound(``int``[] arr,``int` `element)` `  ``{` `    ``for``(``int` `i = 0; i < arr.Length; i++)` `      ``if``(element <= arr[i])` `        ``return` `i;`   `    ``return` `arr.Length;  ` `  ``}`   `  ``// Driver function` `  ``public` `static` `void` `Main (``string``[] args)` `  ``{` `    ``int``[] arr = { 1, 9 };` `    ``int``[] Q = { 7, 10, 0 };`   `    ``minimumIndex(arr, Q);` `  ``}` `}`   `// This code is contributed by phasing17`

## Javascript

 ``

Output:

`1 -1 0`

Time Complexity: O(N * log(N))
Auxiliary Space: O(N)

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