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:

  1. Create an auxiliary array nextpos[] to store for every array element, the index of the next element which is different from the current element.
  2. 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.
  3. 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].
  4. If nextpos[L] is less than equal to R, then print nexpos[L].
  5. 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++

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// 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;
}

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Java

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// 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

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Output:

1
2
-1

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

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Improved By : amit143katiyar