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 ith index, update minIdx[i] = min(minIdx[i + 1], storeArrIdx[i].second).
- Traverse the array Q[] using variable i. For every ith 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 Codeint main(){ vector<int> arr = { 1, 9 }; vector<int> Q = { 7, 10, 0 }; minimumIndex(arr, Q); return 0;} |
Java
// Java program for above approachimport 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 approachffrom 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 Codeif __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 approachusing 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 approachclass 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 functionlet 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)
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