Sudo Placement[1.5] | Second Smallest in Range

Given an array of N integers and Q queries. Every query consists of L and R. The task is to print the second smallest element in range L-R. Print -1 if no second smallest element exists.

Examples:

Input:
a[] = {1, 2, 2, 4}
Queries= 2
L = 1, R = 2
L = 0, R = 1

Output:
-1
2

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Process every query and print the second smallest using the following algorithm.

Initialize both first and second smallest as INT_MAX
Loop through all the elements.
1. If the current element is smaller than first, then update first and second.
2. Else if the current element is smaller than second then update second

If the second element is still INT_MAX after looping through all the elements, print -1 else print the second smallest element.

Below is the implementation of the above approach:

C++

// C++ program for 
// SP - Second Smallest in Range 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to find the second smallest element 
// in range L-R of an array 
int secondSmallest(int a[], int n, int l, int r) 
{ 
  
    int first = INT_MAX; 
    int second = INT_MAX; 
    for (int i = l; i <= r; i++) { 
        if (a[i] < first) { 
            second = first; 
            first = a[i]; 
        } 
        else if (a[i] < second and a[i] != first) { 
            second = a[i]; 
        } 
    } 
  
    if (second == INT_MAX) 
        return -1; 
    else
        return second; 
} 
  
// function to perform queries 
void performQueries(int a[], int n) 
{ 
    // 1-st query 
    int l = 1; 
    int r = 2; 
    cout << secondSmallest(a, n, l, r) << endl; 
  
    // 2nd query 
    l = 0; 
    r = 1; 
    cout << secondSmallest(a, n, l, r); 
} 
  
// Driver Code 
int main() 
{ 
    int a[] = { 1, 2, 2, 4 }; 
    int n = sizeof(a) / sizeof(a[0]); 
    performQueries(a, n); 
    return 0; 
} 

Java

// Java program for 
// SP - Second Smallest in Range 
class GFG  
{ 
// Function to find the  
// second smallest element 
// in range L-R of an array 
static int secondSmallest(int a[], int n,  
                          int l, int r) 
{ 
int first = Integer.MAX_VALUE; 
int second = Integer.MAX_VALUE; 
for (int i = l; i <= r; i++)  
{ 
    if (a[i] < first)  
    { 
        second = first; 
        first = a[i]; 
    } 
    else if (a[i] < second &&  
             a[i] != first)  
    { 
        second = a[i]; 
    } 
} 
  
if (second == Integer.MAX_VALUE) 
    return -1; 
else
    return second; 
} 
  
// function to perform queries 
static void performQueries(int a[], int n) 
{ 
    // 1-st query 
    int l = 1; 
    int r = 2; 
    System.out.println(secondSmallest(a, n, l, r)); 
      
    // 2nd query 
    l = 0; 
    r = 1; 
    System.out.println(secondSmallest(a, n, l, r)); 
} 
  
// Driver Code 
public static void main(String[] args) 
{ 
    int a[] = { 1, 2, 2, 4 }; 
    int n = a.length; 
    performQueries(a, n); 
} 
} 
  
// This code is contributed  
// by ChitraNayal 

Python

# Python program for 
# SP - Second Smallest in Range 
  
# Function to find the  
# second smallest element 
# in range L-R of an array 
import sys 
def secondSmallest(a, n, l, r): 
  
    first = sys.maxsize 
    second = sys.maxsize 
    for i in range(l, r + 1):  
      
        if (a[i] < first):  
          
            second = first 
            first = a[i] 
          
        elif (a[i] < second and 
              a[i] != first):  
          
            second = a[i] 
  
    if (second == sys.maxsize): 
        return -1
    else: 
        return second 
  
  
# function to perform queries 
def performQueries(a, n): 
  
    # 1-st query 
    l = 1
    r = 2
    print(secondSmallest(a, n, l, r))  
  
    # 2nd query 
    l = 0
    r = 1
    print(secondSmallest(a, n, l, r)) 
  
# Driver Code 
a = [1, 2, 2, 4 ] 
n = len(a)  
performQueries(a, n); 
      
# This code is contributed 
# by Shivi_Aggarwal 
     

C#

// C# program for 
// SP - Second Smallest in Range 
using System; 
  
class GFG  
{ 
// Function to find the  
// second smallest element 
// in range L-R of an array 
static int secondSmallest(int[] a, int n,  
                        int l, int r) 
{ 
  
int first = int.MaxValue; 
int second = int.MaxValue; 
for (int i = l; i <= r; i++) 
{ 
    if (a[i] < first) 
    { 
        second = first; 
        first = a[i]; 
    } 
    else if (a[i] < second && 
            a[i] != first)  
    { 
        second = a[i]; 
    } 
} 
  
if (second == int.MaxValue) 
    return -1; 
else
    return second; 
} 
  
// function to perform queries 
static void performQueries(int[] a, int n) 
{ 
    // 1-st query 
    int l = 1; 
    int r = 2; 
    Console.WriteLine(secondSmallest(a, n, l, r)); 
  
    // 2nd query 
    l = 0; 
    r = 1; 
    Console.WriteLine(secondSmallest(a, n, l, r)); 
} 
  
// Driver Code 
public static void Main() 
{ 
    int[] a = { 1, 2, 2, 4 }; 
    int n = a.Length; 
    performQueries(a, n); 
} 
} 
  
// This code is contributed 
// by ChitraNayal 

PHP

<?php  
// PHP program for 
// SP - Second Smallest in Range 
  
// Function to find the  
// second smallest element 
// in range L-R of an array 
function secondSmallest(&$a, $n, $l, $r) 
{ 
    $first = PHP_INT_MAX; 
    $second = PHP_INT_MAX; 
    for ($i = $l; $i <= $r; $i++) 
    { 
        if ($a[$i] < $first)  
        { 
            $second = $first; 
            $first = $a[$i]; 
        } 
        else if ($a[$i] < $second and 
                 $a[$i] != $first) 
        { 
            $second = $a[$i]; 
        } 
    } 
  
    if ($second == PHP_INT_MAX) 
        return -1; 
    else
        return $second; 
} 
  
// function to perform queries 
function performQueries(&$a, $n) 
{ 
    // 1-st query 
    $l = 1; 
    $r = 2; 
    echo secondSmallest($a, $n,  
                        $l, $r)."\n"; 
  
    // 2nd query 
    $l = 0; 
    $r = 1; 
    echo secondSmallest($a, $n,  
                        $l, $r)."\n"; 
} 
  
// Driver Code 
$a = array(1, 2, 2, 4); 
$n = sizeof($a); 
performQueries($a, $n); 
  
// This code is contributed 
// by ChitraNayal 
?> 

Output:

-1
2

Time Complexity: O(M), where M = R-L is the number of elements in range [L, R]

Note: Since the constraints of the question was very less, a brute force solution will pass. The solution can be further optimised using Segment Tree.

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python

C#

PHP

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