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:

• Initialize an array, say storeArrIdx[], of the form { first, second } to store the array elements along with the index.
• Sort the array storeArridx[] in increasing order of the array elements.
• Sort the array arr[] in increasing order.
• Initialize an array, say minIdx[], such that minIdx[i] store the smallest index of an array element whose value is greater than or equal to arr[i].
• Traverse the array storeArrIdx[] in reverse using variable i. For every i^th index, update minIdx[i] = min(minIdx[i + 1], storeArrIdx[i].second).
• Traverse the array Q[] using variable i. For every i^th query, find the index of lower_bound of Q[i] into arr[] and check if the obtained index is less than N or not. If found to be true, then print minIdx[i].
• Otherwise, print -1.

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach
#include <bits/stdc++.h>
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<pair<int, int> > 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<pair> 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<pair> storeArrIdx = new List<pair>();

    // 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

<script>

// JavaScript program for above approach

class pair
{
    constructor(element,index)
    {
        this.element = element;
        this.index = index;
    }
}

// Function to find the smallest index
  // of an array element whose value is
  // less than or equal to Q[i]
function minimumIndex(arr,Q)
{
    // Stores size of array
    let N = arr.length;

    // Stores count of queries
    let M = Q.length;

    // Store array elements along
    // with the index
    let storeArrIdx = [];

    // Store smallest index of an array
    // element whose value is greater
    // than or equal to i
    let minIdx = new Array(N);
     for(let i=0;i<N;i++)
    {
        minIdx[i]=0;
    }
    // Traverse the array
    for (let i = 0; i < N; ++i)
    {

      // Insert {arr[i], i} into
      // storeArrIdx[]
      storeArrIdx.push([arr[i], i]);
    }

    // Sort the array
    (arr).sort(function(a,b){return a-b;});

    // Sort the storeArrIdx
    storeArrIdx.sort(function(a, b){ return a[0]-b[0]});

    // Stores index of arr[N - 1] in
    // sorted order
    minIdx[N - 1]
      = storeArrIdx[N - 1][1];

    // Traverse the array storeArrIdx[]
    for (let i = N - 2; i >= 0; i--) {

      // Update minIdx[i]
      minIdx[i] =Math.min(minIdx[i + 1],
                          storeArrIdx[i][1]);
    }

    // Traverse the array Q[]
    for (let i = 0; i < M; i++) {

      // Store the index of
      // lower_bound of Q[i]
      let pos
        = lower_bound(arr, Q[i]);

      // If no index found whose value
      // greater than or equal to arr[i]
      if (pos == N) {
        document.write("-1"+" ");
        continue;
      }

      // Print smallest index whose value
      // greater than or equal to Q[i]
      document.write(minIdx[pos]+" ");
    }
}

function lower_bound(arr,element)
{
    for(let i = 0; i < arr.length; i++)
      if(element <= arr[i])
        return i;

    return arr.length;
}

// Driver function
let arr=[1, 9];
let Q=[7, 10, 0 ];

minimumIndex(arr, Q);


// This code is contributed by patel2127

</script>

Output:

1 -1 0

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