Sudo Placement[1.5] | Second Smallest in Range
Last Updated :
11 Jul, 2025
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
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.
- If the current element is smaller than the first, then update first and second.
- Else if the current element is smaller than the 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 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 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# 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 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
?>
<script>
// Javascript 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)
{
let first = Number.MAX_VALUE;
let second = Number.MAX_VALUE;
for (let 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 == Number.MAX_VALUE)
return -1;
else
return second;
}
// function to perform queries
function performQueries(a,n)
{
// 1-st query
let l = 1;
let r = 2;
document.write(secondSmallest(a, n, l, r)+"<br>");
// 2nd query
l = 0;
r = 1;
document.write(secondSmallest(a, n, l, r)+"<br>");
}
// Driver Code
let a=[1, 2, 2, 4];
let n = a.length;
performQueries(a, n);
// This code is contributed by rag2127
</script>
Time Complexity: O(M), where M = R-L is the number of elements in the range [L, R]
Auxiliary Space: O(1)
Note: Since the constraints of the question were very less, a brute force solution will pass. The solution can be further optimized using a Segment Tree.
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