Check if array contains contiguous integers with duplicates allowed

Given an array Arr of N integers(duplicates allowed). Print “Yes” if it is a set of contiguous integers else print “No”.

Examples:

Input: Arr = { 5, 2, 3, 6, 4, 4, 6, 6 }
Output: Yes
The elements of array form a contiguous set of integers which is {2, 3, 4, 5, 6} so the output is Yes.



Input: Arr = { 10, 14, 10, 12, 12, 13, 15 }
Output: No

Approach:

  1. A similar approach is discussed here. It is recommonded to go through it before moving further with this article.
  2. Find the max and min elements from the array.
  3. Update the array with the difference from the max element in the array.
  4. Traverse the array and do the following
    • If 0 <= abs(arr[i])-1) < n and arr[abs(arr[i])-1)] > 0 means the element is positive (visited first time), make it negative by doing arr[abs(arr[i])-1] = -arr[abs(arr[i])-1].
    • Else continue loop means either the value is already visited or out of range of the array
  5. Traverse the loop again and check if any element is positive means element has a difference greater than 1
    break the loop and print “NO”
  6. If no element is found positive, print “YES”

Below is the implementation code:

C++

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// C++ program to check if
// array contains contiguous
// integers with duplicates
// allowed in O(1) space
#include <bits/stdc++.h>
using namespace std;
  
// Function to return true or
// false
bool check(int arr[], int n)
{
  
    int k = INT_MIN;
    int r = INT_MAX;
  
    // To find the max and min
    // in the array
    for (int i = 0; i < n; i++) {
        k = max(k, arr[i]);
        r = min(r, arr[i]);
    }
  
    k += 1;
  
    // Update the array with
    // the difference from
    // the max element
    for (int i = 0; i < n; i++)
        arr[i] = k - arr[i];
  
    for (int i = 0; i < n; i++) {
  
        // if the element is positive
        // and less than the size of
        // array(in range), make it negative
        if (abs(arr[i]) - 1 < n
            && arr[abs(arr[i]) - 1] > 0) {
            arr[abs(arr[i]) - 1]
                = -arr[abs(arr[i]) - 1];
        }
    }
  
    int flag = 0;
  
    // Loop from 0 to end of the array
    for (int i = 0; i <= k - r - 1; i++) {
  
        // Found positive, out of range
        if (arr[i] > 0) {
            flag = 1;
            break;
        }
    }
  
    return flag == 0;
}
  
// Driver function
int main()
{
  
    // Given array
    int arr[] = { 5, 2, 3, 6, 4, 4, 6, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    // Function Calling
    if (check(arr, n))
        cout << "Yes";
    else
        cout << "No";
  
    return 0;
}

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Python3

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# Python3 program to check if array contains 
# contiguous integers with duplicates allowed 
# in O(1) space
  
# Function to return true or false
def check(arr, n):
  
    k = -10**9
    r = 10**9
  
    # To find the max and min in the array
    for i in range(n):
        k = max(k, arr[i])
        r = min(r, arr[i])
  
    k += 1
  
    # Update the array with the difference 
    # from the max element
    for i in range(n):
        arr[i] = k - arr[i]
  
    for i in range(n):
  
        # if the element is positive
        # and less than the size of
        # array(in range), make it negative
        if (abs(arr[i]) - 1 < n and 
                arr[abs(arr[i]) - 1] > 0):
            arr[abs(arr[i]) - 1]= -arr[abs(arr[i]) - 1]
  
    flag = 0
  
    # Loop from 0 to end of the array
    for i in range(k - r):
  
        # Found positive, out of range
        if (arr[i] > 0):
            flag = 1
            break
  
    return flag == 0
  
# Driver Code
  
# Given array
arr = [5, 2, 3, 6, 4, 4, 6, 6]
n = len(arr)
  
# Function Calling
if (check(arr, n)):
    print("Yes")
else:
    print("No")
  
# This code is contributed by Mohit Kumar

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Output:

Yes

Time Complexity: O(N)
Space Complexity: O(1) Extra Space



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