Largest Element

Intuition

Imagine being a teacher and having a list of marks obtained by your students in a recent test. The task is to find out which student scored the highest marks. One simple way to do this is by sorting the marks in ascending order and then taking the last element from the sorted list.

Approach

Sort the array in ascending order using in-built methods.

Return the element at the last index of the array.

Solution

#include <bits/stdc++.h>
using namespace std;

class Solution {
public:
    static int largestElement(std::vector<int>& nums) {
        // Sort the array
        std::sort(nums.begin(), nums.end());

        // Largest element will be at the last index of the array.
        int largest = nums[nums.size() - 1];

        // Return the largest element in array.
        return largest;
    }
};

int main() {
    std::vector<int> nums = {3, 2, 1, 5, 2};

    // Create an instance of the Solution class
    Solution sol;

    int largest = sol.largestElement(nums);

    // Print the largest element.
    std::cout << largest << std::endl;

    return 0;
}
import java.util.*;

class Solution {
    public static int largestElement(int[]  nums) {
        // Sort array
        Arrays.sort(nums);

        /*Largest element will be at 
        the last index of the array.*/
        int largest = nums[nums.length - 1];

        //Return the largest element in array.
        return largest;
    }

    public static void main(String[] args) {
        int[] nums = {3,2,1,5,2};

        //Make an intance of the Solution class
        Solution sol = new Solution();

        int largest = sol.largestElement(nums);

        // Print the largest element.
        System.out.print(largest);
    }
}
from typing import List

class Solution:
    def largestElement(self, nums):
        # Sort the list 
        nums.sort()

        # Largest element will be 
        # at the last index of the list
        largest = nums[-1]

        # Return the largest element
        return largest

# Main function
if __name__ == "__main__":
    nums = [3, 2, 1, 5, 2]

# Create an instance of the Solution class
    sol = Solution()
    
    largest = sol.largestElement(nums)

    # Print the largest element
    print(largest)

class Solution {

    largestElement(nums) {
        // Sort array
        nums.sort((a, b) => a - b);

        /** The largest element will be at the 
           last index of the array*/
        const largest = nums[nums.length - 1];

        // Return the largest element
        return largest;
    }
}

const nums = [3, 2, 1, 5, 2];

// Create an instance of the Solution class
const sol = new Solution();

const largest = sol.largestElement(nums);

// Print the largest element
console.log(largest);

Complexity Analysis

Time Complexity: O(N * logN), as we are sorting the array, where N is the length of the array. Space Complexity: O(1), since no extra space is used.

Intuition

Imagine being a teacher having a list of marks obtained by your students in a recent test. The ask is to find out which student scored the highest marks.

Now to find the highest marks in the list, keep track of the marks encountered so far in a variable, let's say maxi. Starting value of maxi will be the first marksheet's marks as initially that will the maximum.

Then go checking each marksheet one by one and keep comparing and updating maxi as and when a marks greater than current marks stored in maxi is found. This will ensure that once we have all the marksheets checked we will have maximum marks stored in the variable maxi.

Approach

Create a max variable and initialize it with arr[0], as in the beginning the first element should be maximum.

Iterate in the array and compare it with other elements to update the maximum.

If any element is greater than the max value, update max value with the element’s value and at last return the max.

Dry Run:

Solution

#include <bits/stdc++.h>
using namespace std;

class Solution {
public:
    int largestElement(vector<int>& nums) {
        // Initialize max as the first element
        int max = nums[0];

        // Traverse the entire array
        for (int i = 1; i < nums.size(); i++) {

            /* If current element is greater
            than max, update max*/
            if (nums[i] > max) {
                max = nums[i];
            }

        }
        // Return the largest element found
        return max;
    }
};

int main() {
    vector nums = {3, 2, 1, 5, 2};

 // Create an instance of the Solution class
    Solution sol;

    int largest = sol.largestElement(nums);

    // Print the largest element in the array.
    cout << largest << endl;
    return 0;
}
import java.util.*;

class Solution {
    public int largestElement(int[] nums) {
        // Initialize max as the first element
        int max = nums[0];

        // Traverse the entire array
        for (int i = 1; i < nums.length; i++) {

            /* If current element is greater
            than max, update max*/
            if (nums[i] > max) {
                max = nums[i];
            }

        }
        // Return the largest element found
        return max;
    }

    public static void main(String[] args) {
        int[] nums = {3,2,1,5,2};

        // Create an instance of the Solution class
        Solution sol = new Solution();

        // Find and print the largest element in the array
        System.out.println("The largest element in the array is: " + sol.largestElement(nums));
    }
}
from typing import List

class Solution:
    def largestElement(self, nums):
        # Initialize max_val as the first element
        max_val = nums[0]

        # Traverse the entire list
        for num in nums[1:]:

            """
            If the current element is greater 
            than max_val, update max_val
            """
            if num > max_val:
                max_val = num

        # Return the largest element found
        return max_val

# Main function
if __name__ == "__main__":
    nums = [3, 2, 1, 5, 2]

    # Create an instance of the Solution class
    sol = Solution()

    # Find and print the largest element in the list
    print("The largest element in the list is:", sol.largestNumber(nums))
class Solution {
    
    largestElement(nums) {
        // Initialize max as the first element
        let max = nums[0];

        // Traverse the entire array
        for (let i = 1; i < nums.length; i++) {

            /*If current element is greater
            than max, update max*/
            if (nums[i] > max) {
                max = nums[i];

            }
        }
        // Return the largest element found
        return max;
    }
}

const nums = [3, 2, 1, 5, 2];

// Create an instance of the Solution class
const sol = new Solution();

// Find and print the largest element in the array
console.log("The largest element in the array is: " + sol.largestElement(nums));

Complexity Analysis

Time Complexity: O(N) , since there is linear traversal of the array, where N is the length of the array.
Space Complexity: O(1), as no additional space is used.

Problem List

Examples:

Constraints

Hints

Company Tags

Editorial

Intuition

Approach

Solution

Complexity Analysis

Intuition

Approach

Dry Run:

Solution

Complexity Analysis

Frequently Occurring Doubts

Interview Followup Questions

Notes

Code