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
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 arrayint 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 queriesvoid 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 Codeint 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 Rangeclass GFG{// Function to find the// second smallest element// in range L-R of an arraystatic 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 queriesstatic 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 Codepublic 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 arrayimport sysdef 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 queriesdef 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 Codea = [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 Rangeusing System;class GFG{// Function to find the// second smallest element// in range L-R of an arraystatic 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 queriesstatic 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 Codepublic 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 arrayfunction 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 $second;}// function to perform queriesfunction performQueries(&$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?> |
Javascript
<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 queriesfunction 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 Codelet a=[1, 2, 2, 4];let n = a.length;performQueries(a, n); // This code is contributed by rag2127</script> |
Output:
-1 2
Time Complexity: O(M), where M = R-L is the number of elements in the range [L, R]
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.
