Search an element in given N ranges
Last Updated :
11 Jul, 2025
Given an array of N sorted ranges and a number K. The task is to find the index of the range in which K lies. If K does not lie in any of the given ranges then print -1.
Note: None of the given ranges coincide.
Examples:
Input: arr[] = { { 1, 3 }, { 4, 7 }, { 8, 11 } }, K = 6
Output: 1
6 lies in the range {4, 7} with index = 1
Input: arr[] = { { 1, 3 }, { 4, 7 }, { 9, 11 } }, K = 8
Output: -1
Naive approach: The following steps can be followed to solve the above problem.
- Traverse all the ranges.
- Check if the condition K >= arr[i].first && K <= arr[i].second holds in any of the iterations.
- If the number K does not lie in any of the given ranges then print -1.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the index of the range
// in which K lies and uses linear search
int findNumber(pair<int, int> a[], int n, int K)
{
// Iterate and find the element
for (int i = 0; i < n; i++) {
// If K lies in the current range
if (K >= a[i].first && K <= a[i].second)
return i;
}
// K doesn't lie in any of the given ranges
return -1;
}
// Driver code
int main()
{
pair<int, int> a[] = { { 1, 3 }, { 4, 7 }, { 8, 11 } };
int n = sizeof(a) / sizeof(a[0]);
int k = 6;
int index = findNumber(a, n, k);
if (index != -1)
cout << index;
else
cout << -1;
return 0;
}
// Java implementation of the approach
class GFG
{
static class pair
{
int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to return the index
// of the range in which K lies
// and uses linear search
static int findNumber(pair a[],
int n, int K)
{
// Iterate and find the element
for (int i = 0; i < n; i++)
{
// If K lies in the current range
if (K >= a[i].first &&
K <= a[i].second)
return i;
}
// K doesn't lie in any
// of the given ranges
return -1;
}
// Driver code
public static void main(String[] args)
{
pair a[] = {new pair(1, 3 ),
new pair(4, 7 ),
new pair(8, 11 )};
int n = a.length;
int k = 6;
int index = findNumber(a, n, k);
if (index != -1)
System.out.println(index);
else
System.out.println(-1);
}
}
// This code is contributed by Rajput-Ji
# Python 3 implementation of the approach
# Function to return the index of the range
# in which K lies and uses linear search
def findNumber(a, n, K):
# Iterate and find the element
for i in range(0, n, 1):
# If K lies in the current range
if (K >= a[i][0] and K <= a[i][1]):
return i
# K doesn't lie in any of the
# given ranges
return -1
# Driver code
if __name__ == '__main__':
a = [[1, 3], [4, 7], [8, 11]]
n = len(a)
k = 6
index = findNumber(a, n, k)
if (index != -1):
print(index, end = "")
else:
print(-1, end = "")
# This code is contributed by
# Surendra_Gangwar
// C# implementation of the approach
using System;
class GFG
{
class pair
{
public int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to return the index
// of the range in which K lies
// and uses linear search
static int findNumber(pair []a,
int n, int K)
{
// Iterate and find the element
for (int i = 0; i < n; i++)
{
// If K lies in the current range
if (K >= a[i].first &&
K <= a[i].second)
return i;
}
// K doesn't lie in any
// of the given ranges
return -1;
}
// Driver code
public static void Main(String[] args)
{
pair []a = {new pair(1, 3 ),
new pair(4, 7 ),
new pair(8, 11 )};
int n = a.Length;
int k = 6;
int index = findNumber(a, n, k);
if (index != -1)
Console.WriteLine(index);
else
Console.WriteLine(-1);
}
}
// This code is contributed by 29AjayKumar
<script>
// JavaScript implementation of the approach
// Function to return the index of the range
// in which K lies and uses linear search
function findNumber(a, n, K)
{
// Iterate and find the element
for (var i = 0; i < n; i++) {
// If K lies in the current range
if (K >= a[i][0] && K <= a[i][1])
return i;
}
// K doesn't lie in any of the given ranges
return -1;
}
// Driver code
var a = [ [ 1, 3 ], [ 4, 7 ], [ 8, 11 ] ];
var n = a.length;
var k = 6;
var index = findNumber(a, n, k);
if (index != -1)
document.write( index);
else
document.write( -1);
</script>
Time Complexity: O(N), as we are using a loop to traverse N times to check if the number lies in the given range. Where N is the number of pairs in the array.
Auxiliary Space: O(1), as we are not using any extra space.
Efficient Approach: Binary Search can be used to find the element in O(log N).
Below is the implementation of the above approach:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the index of the range
// in which K lies and uses binary search
int findNumber(pair<int, int> a[], int n, int K)
{
int low = 0, high = n - 1;
// Binary search
while (low <= high) {
// Find the mid element
int mid = (low + high) >> 1;
// If element is found
if (K >= a[mid].first && K <= a[mid].second)
return mid;
// Check in first half
else if (K < a[mid].first)
high = mid - 1;
// Check in second half
else
low = mid + 1;
}
// Not found
return -1;
}
// Driver code
int main()
{
pair<int, int> a[] = { { 1, 3 }, { 4, 7 }, { 8, 11 } };
int n = sizeof(a) / sizeof(a[0]);
int k = 6;
int index = findNumber(a, n, k);
if (index != -1)
cout << index;
else
cout << -1;
return 0;
}
// Java implementation of the approach
class GFG
{
static class pair
{
int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to return the index of the range
// in which K lies and uses binary search
static int findNumber(pair a[], int n, int K)
{
int low = 0, high = n - 1;
// Binary search
while (low <= high)
{
// Find the mid element
int mid = (low + high) >> 1;
// If element is found
if (K >= a[mid].first &&
K <= a[mid].second)
return mid;
// Check in first half
else if (K < a[mid].first)
high = mid - 1;
// Check in second half
else
low = mid + 1;
}
// Not found
return -1;
}
// Driver code
public static void main(String[] args)
{
pair a[] = { new pair(1, 3),
new pair(4, 7),
new pair(8, 11) };
int n = a.length;
int k = 6;
int index = findNumber(a, n, k);
if (index != -1)
System.out.println(index);
else
System.out.println(-1);
}
}
// This code is contributed by Princi Singh
# Python3 implementation of the approach
# Function to return the index of the range
# in which K lies and uses binary search
def findNumber(a, n, K):
low = 0
high = n - 1
# Binary search
while (low <= high):
# Find the mid element
mid = (low + high) >> 1
# If element is found
if (K >= a[mid][0] and K <= a[mid][1]):
return mid
# Check in first half
elif (K < a[mid][0]):
high = mid - 1
# Check in second half
else:
low = mid + 1
# Not found
return -1
# Driver code
a= [ [ 1, 3 ], [ 4, 7 ], [ 8, 11 ] ]
n = len(a)
k = 6
index = findNumber(a, n, k)
if (index != -1):
print(index)
else:
print(-1)
# This code is contributed by mohit kumar
// C# implementation of the above approach
using System;
class GFG
{
public class pair
{
public int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to return the index of the range
// in which K lies and uses binary search
static int findNumber(pair []a, int n, int K)
{
int low = 0, high = n - 1;
// Binary search
while (low <= high)
{
// Find the mid element
int mid = (low + high) >> 1;
// If element is found
if (K >= a[mid].first &&
K <= a[mid].second)
return mid;
// Check in first half
else if (K < a[mid].first)
high = mid - 1;
// Check in second half
else
low = mid + 1;
}
// Not found
return -1;
}
// Driver code
public static void Main(String[] args)
{
pair []a = {new pair(1, 3),
new pair(4, 7),
new pair(8, 11)};
int n = a.Length;
int k = 6;
int index = findNumber(a, n, k);
if (index != -1)
Console.WriteLine(index);
else
Console.WriteLine(-1);
}
}
// This code is contributed by Rajput-Ji
<script>
// JavaScript implementation of the approach
class pair {
constructor(first , second) {
this.first = first;
this.second = second;
}
}
// Function to return the index of the range
// in which K lies and uses binary search
function findNumber( a , n , K) {
var low = 0, high = n - 1;
// Binary search
while (low <= high) {
// Find the mid element
var mid = (low + high) >> 1;
// If element is found
if (K >= a[mid].first && K <= a[mid].second)
return mid;
// Check in first half
else if (K < a[mid].first)
high = mid - 1;
// Check in second half
else
low = mid + 1;
}
// Not found
return -1;
}
// Driver code
var a = [ new pair(1, 3), new pair(4, 7),
new pair(8, 11) ];
var n = a.length;
var k = 6;
var index = findNumber(a, n, k);
if (index != -1)
document.write(index);
else
document.write(-1);
// This code contributed by gauravrajput1
</script>
Time Complexity: O(log N), as we are using binary search, in each traversal we divide the array into two halves and choose one of them to search further in. So the effective time is 1+1/2+1/4+....+1/2^N which is equivalent to logN. Where N is the number of pairs in the array.
Auxiliary Space: O(1), as we are not using any extra space.
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