Find the largest after deleting the given elements

Given an array of integers, find the largest number after deleting the given elements. In case of repeating elements, delete one instance for every instance of the element present in the array containing the elements to be deleted.

Examples:

Input :array[] = { 5, 12, 33, 4, 56, 12, 20 }
del[] = { 12, 33, 56, 5 }
Output : 20
Explanation : We get {12, 20} after deleting given elements. Largest among remaining element is 20

Approach :

  • Insert all the numbers in the hash map which are to be deleted from the array, so that we can check if the element in the array is also present in the Delete-array in O(1) time.
  • Initialize largest number max to be INT_MIN.
  • Traverse through the array. Check if the element is present in the hash map.
  • If present, erase it from the hash map, else if not present compare it with max variable and change its value if the value of the element is greater than the max value.

C++

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// C++ program to find the largest number
// from the array after  n deletions
#include "climits"
#include "iostream"
#include "unordered_map"
using namespace std;
  
// Returns maximum element from arr[0..m-1] after deleting
// elements from del[0..n-1]
int findlargestAfterDel(int arr[], int m, int del[], int n)
{
    // Hash Map of the numbers to be deleted
    unordered_map<int, int> mp;
    for (int i = 0; i < n; ++i) {
  
        // Increment the count of del[i]
        mp[del[i]]++;
    }
  
    // Initializing the largestElement
    int largestElement = INT_MIN;
  
    for (int i = 0; i < m; ++i) {
  
        // Search if the element is present
        if (mp.find(arr[i]) != mp.end()) {
  
            // Decrement its frequency
            mp[arr[i]]--;
  
            // If the frequency becomes 0,
            // erase it from the map
            if (mp[arr[i]] == 0)
                mp.erase(arr[i]);
        }
  
        // Else compare it largestElement
        else
            largestElement = max(largestElement, arr[i]);
    }
  
    return largestElement;
}
  
int main()
{
    int array[] = { 5, 12, 33, 4, 56, 12, 20 };
    int m = sizeof(array) / sizeof(array[0]);
  
    int del[] = { 12, 33, 56, 5 };
    int n = sizeof(del) / sizeof(del[0]);
  
    cout << findlargestAfterDel(array, m, del, n);
    return 0;
}

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Python3

# Python3 program to find the largest
# number from the array after n deletions
import math as mt

# Returns maximum element from arr[0..m-1]
# after deleting elements from del[0..n-1]
def findlargestAfterDel(arr, m, dell, n):

# Hash Map of the numbers
# to be deleted
mp = dict()
for i in range(n):

# Increment the count of del[i]
if dell[i] in mp.keys():
mp[dell[i]] += 1
else:
mp[dell[i]] = 1

# Initializing the largestElement
largestElement = -10**9

for i in range(m):

# Search if the element is present
if (arr[i] in mp.keys()):

# Decrement its frequency
mp[arr[i]] -= 1

# If the frequency becomes 0,
# erase it from the map
if (mp[arr[i]] == 0):
mp.pop(arr[i])

# Else compare it largestElement
else:
largestElement = max(largestElement,
arr[i])

return largestElement

# Driver code
array = [5, 12, 33, 4, 56, 12, 20]
m = len(array)

dell = [12, 33, 56, 5]
n = len(dell)

print(findlargestAfterDel(array, m, dell, n))

# This code is contributed
# by mohit kumar 29

Output :

20

Time Complexity – O(N)



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