Smallest index in given range of indices which is not equal to X
Given an integer array arr[] of size N, and Q queries of the form of {L, R, X}, the task is to find the smallest index between L and R from the given array such that arr[i] != X. If no such index exists in array, print -1.
Examples:
Input: arr[] = {1, 2, 1, 1, 3, 5}, query[][] = {{0, 3, 1}, {1, 5, 2}, {2, 3, 1}}
Output: 1
2
-1
Explanation:
The first index from 0 to 3 which does not contain 1 is 1
The first index from 1 to 5 which does not contain 2 is 2
All the indices from 2 to 3 contain 1. Hence, the answer is -1.Input: arr[] = { 1, 2, 3, 2, 1, 1, 3, 5}, query[][] = { { 1, 4, 2 }, { 4, 5, 1 }, { 2, 3, 1 }, { 4, 6, 1 }}
Output: 2
-1
2
6
Naive Approach:
Iterate over the range [L, R] for every query and check if any index exists which does not contain X . If such an index is found, print that index. Otherwise, print -1.
Time Complexity: O (N * Q)
Auxiliary Space: O (1)
Efficient Approach:
The above approach can be further optimized by precomputing and storing the index of the next element which is different from the current element, for every array element, which reduces computational complexity to O(1) for every query.
Follow the steps below to solve the problem:
- Create an auxiliary array nextpos[] to store for every array element, the index of the next element which is different from the current element.
- Now to process every query, firstly check if the value at index L is not equal to X. If so, then the answer will be L.
- Otherwise, it means that arr[L] = X. In this case, we need to find the next index which has a value different from arr[L]. This can be obtained from nextpos[L].
- If nextpos[L] is less than equal to R, then print nexpos[L].
- If neither of the above conditions are satisfied, then answer will be -1.
Illustration:
arr[] = {1, 2, 1, 1, 3, 5}, query[][] = {{0, 3, 1}, {1, 5, 2}, {2, 3, 1}}
For the given arr[], nextpos[] array will be {1, 2, 4, 4, 5, 6}
For 1st Query: L = 0, R = 3, X = 1
arr[0] = 1 = X
nextpos[0] = 1( < 3)
Hence, the answer is 1.
For the 2nd Query: L = 1, R = 5, X = 2
arr[1] = 2( = X)
nextpos[1] = 2( < 5)
Hence, the answer is 2.
For the 3rd Query: L = 2, R = 3, X = 1
arr[2] = 1( = X)
nextpos[2] = 4( > R)
Hence, the answer is -1
Below is the implementation of the above approach:
C++
// C++ Program to find the smallest // index in the array in the range // [L, R] which does not contain X #include <bits/stdc++.h> using namespace std; // Precompute the index of next // different element in the array // for every array element void precompute( int nextpos[], int arr[], int N) { // Default value nextpos[N - 1] = N; for ( int i = N - 2; i >= 0; i--) { // Compute nextpos[i] using // nextpos[i+1] if (arr[i] == arr[i + 1]) nextpos[i] = nextpos[i + 1]; else nextpos[i] = i + 1; } } // Function to return the smallest index void findIndex( int query[][3], int arr[], int N, int Q) { // nextpos[i] will store the next // position p where arr[p]!=arr[i] int nextpos[N]; precompute(nextpos, arr, N); for ( int i = 0; i < Q; i++) { int l, r, x; l = query[i][0]; r = query[i][1]; x = query[i][2]; int ans = -1; // If X is not present at l if (arr[l] != x) ans = l; // Otherwise else { // Find the index which // stores a value different // from X int d = nextpos[l]; // If that index is within // the range if (d <= r) ans = d; } cout << ans << "\n" ; } } // Driver Code int main() { int N, Q; N = 6; Q = 3; int arr[] = { 1, 2, 1, 1, 3, 5 }; int query[Q][3] = { { 0, 3, 1 }, { 1, 5, 2 }, { 2, 3, 1 } }; findIndex(query, arr, N, Q); return 0; } |
Java
// Java program to find the smallest // index in the array in the range // [L, R] which does not contain X class GFG{ // Precompute the index of next // different element in the array // for every array element static void precompute( int nextpos[], int arr[], int N) { // Default value nextpos[N - 1 ] = N; for ( int i = N - 2 ; i >= 0 ; i--) { // Compute nextpos[i] using // nextpos[i+1] if (arr[i] == arr[i + 1 ]) nextpos[i] = nextpos[i + 1 ]; else nextpos[i] = i + 1 ; } } // Function to return the smallest index static void findIndex( int query[][], int arr[], int N, int Q) { // nextpos[i] will store the next // position p where arr[p]!=arr[i] int []nextpos = new int [N]; precompute(nextpos, arr, N); for ( int i = 0 ; i < Q; i++) { int l, r, x; l = query[i][ 0 ]; r = query[i][ 1 ]; x = query[i][ 2 ]; int ans = - 1 ; // If X is not present at l if (arr[l] != x) ans = l; // Otherwise else { // Find the index which // stores a value different // from X int d = nextpos[l]; // If that index is within // the range if (d <= r) ans = d; } System.out.print(ans + "\n" ); } } // Driver Code public static void main(String[] args) { int N, Q; N = 6 ; Q = 3 ; int arr[] = { 1 , 2 , 1 , 1 , 3 , 5 }; int query[][] = { { 0 , 3 , 1 }, { 1 , 5 , 2 }, { 2 , 3 , 1 } }; findIndex(query, arr, N, Q); } } // This code is contributed by Amit Katiyar |
Python3
# Python3 program to find the smallest # index in the array in the range # [L, R] which does not contain X # Precompute the index of next # different element in the array # for every array element def precompute(nextpos, arr, N): # Default value nextpos[N - 1 ] = N for i in range (N - 2 , - 1 , - 1 ): # Compute nextpos[i] using # nextpos[i+1] if arr[i] = = arr[i + 1 ]: nextpos[i] = nextpos[i + 1 ] else : nextpos[i] = i + 1 # Function to return the smallest index def findIndex(query, arr, N, Q): # nextpos[i] will store the next # position p where arr[p]!=arr[i] nextpos = [ 0 ] * N precompute(nextpos, arr, N) for i in range (Q): l = query[i][ 0 ] r = query[i][ 1 ] x = query[i][ 2 ] ans = - 1 # If X is not present at l if arr[l] ! = x: ans = l # Otherwise else : # Find the index which # stores a value different # from X d = nextpos[l] # If that index is within # the range if d < = r: ans = d print (ans) # Driver code N = 6 Q = 3 arr = [ 1 , 2 , 1 , 1 , 3 , 5 ] query = [ [ 0 , 3 , 1 ], [ 1 , 5 , 2 ], [ 2 , 3 , 1 ] ] findIndex(query, arr, N, Q) # This code is contributed by divyeshrabadiya07 |
C#
// C# program to find the smallest // index in the array in the range // [L, R] which does not contain X using System; class GFG{ // Precompute the index of next // different element in the array // for every array element static void precompute( int []nextpos, int []arr, int N) { // Default value nextpos[N - 1] = N; for ( int i = N - 2; i >= 0; i--) { // Compute nextpos[i] using // nextpos[i+1] if (arr[i] == arr[i + 1]) nextpos[i] = nextpos[i + 1]; else nextpos[i] = i + 1; } } // Function to return the smallest index static void findIndex( int [,]query, int []arr, int N, int Q) { // nextpos[i] will store the next // position p where arr[p]!=arr[i] int []nextpos = new int [N]; precompute(nextpos, arr, N); for ( int i = 0; i < Q; i++) { int l, r, x; l = query[i, 0]; r = query[i, 1]; x = query[i, 2]; int ans = -1; // If X is not present at l if (arr[l] != x) ans = l; // Otherwise else { // Find the index which // stores a value different // from X int d = nextpos[l]; // If that index is within // the range if (d <= r) ans = d; } Console.Write(ans + "\n" ); } } // Driver Code public static void Main(String[] args) { int N, Q; N = 6; Q = 3; int []arr = { 1, 2, 1, 1, 3, 5 }; int [,]query = { { 0, 3, 1 }, { 1, 5, 2 }, { 2, 3, 1 } }; findIndex(query, arr, N, Q); } } // This code is contributed by Amit Katiyar |
Javascript
<script> // Javascript Program to find the smallest // index in the array in the range // [L, R] which does not contain X // Precompute the index of next // different element in the array // for every array element function precompute(nextpos, arr, N) { // Default value nextpos[N - 1] = N; for ( var i = N - 2; i >= 0; i--) { // Compute nextpos[i] using // nextpos[i+1] if (arr[i] == arr[i + 1]) nextpos[i] = nextpos[i + 1]; else nextpos[i] = i + 1; } } // Function to return the smallest index function findIndex(query, arr, N, Q) { // nextpos[i] will store the next // position p where arr[p]!=arr[i] var nextpos = Array(N); precompute(nextpos, arr, N); for ( var i = 0; i < Q; i++) { var l, r, x; l = query[i][0]; r = query[i][1]; x = query[i][2]; var ans = -1; // If X is not present at l if (arr[l] != x) ans = l; // Otherwise else { // Find the index which // stores a value different // from X var d = nextpos[l]; // If that index is within // the range if (d <= r) ans = d; } document.write( ans + "<br>" ); } } // Driver Code var N, Q; N = 6; Q = 3; var arr = [1, 2, 1, 1, 3, 5]; var query = [ [ 0, 3, 1 ], [ 1, 5, 2 ], [ 2, 3, 1 ] ]; findIndex(query, arr, N, Q); </script> |
1 2 -1
Time Complexity: O(N+Q)
Auxiliary Space: O(N)
Contact Us