Day 21 - Array Series Part 4

Question 1 - Pair Sum N

Question – Given a sorted array of integers and a number num, write a function pairSumN which returns the first pair of integers whose sum equals to the given number num, return an empty array if the pair does not exist

Example

input: arr = [-3, -2, -1, 0, 1, 2, 3], num = 0
output: [-3, 3]

input: arr = [-1, 4, 5, 5, 6, 7, 8], num = 10
output: [4, 6]

input; arr = [1, 2, 3, 4, 5, 6], num = 20
output: []

Question 2 - Max Subarray Sum

Question – Given an array of integers and a number n, write a function which returns the maximum sum of n consecutive numbers in that array.

Example

input: arr = [1, 3, 2, 4, 7, 5, 4], N = 3
output: 16 (7 + 5 + 4)

Question 1 – Pair Sum N

JavaScript Implementation

Solution 1 by @MadhavBahl

/**
 * @author MadhavBahl
 * @date 17/01/2019
 * Method used -- bruteforce search (Iteration using nested loops)
 */

function pairSumN (arr, num) {
    let pairArray = [];

    for (let i=0; i<arr.length-1; i++)
        for (let j=i+1; j<arr.length; j++) {
            if (arr[i] + arr[j] === num) {
                pairArray.push (arr[i], arr[j]);
                return pairArray;
            }
        }
    
    return pairArray;
}

console.log (pairSumN ([-3, -2, -1, 0, 1, 2, 3], 0));
console.log (pairSumN ([-1, 4, 5, 5, 6, 7, 8], 10));
console.log (pairSumN ([1, 2, 3, 4, 5, 6], 20));

Solution 2 by @MadhavBahl

/**
 * @author MadhavBahl
 * @date 16/01/2019
 * Method -- Multiple Pointers, solution in O(n)
 */

function pairSumN (arr, num) {
    let i=0, j=arr.length-1, pairArr = [];

    while (i<j) {
        if (arr[i] + arr[j] === num) {
            pairArr.push (arr[i], arr[j]);
            return pairArr;
        }
        else if (arr[i] + arr[j] < num)
            i++;
        else
            j--;
    }

    return pairArr;
}

console.log (pairSumN ([-3, -2, -1, 0, 1, 2, 3], 0));
console.log (pairSumN ([-1, 4, 5, 5, 6, 7, 8], 10));
console.log (pairSumN ([1, 2, 3, 4, 5, 6], 20));

Ruby Implementation

Solution

=begin
@author: aaditkamat
@date: 17/01/2019
=end
def pair_sum(arr, num)
  map = {}
  arr.size.times do |i|
    map[arr[i]] = num - arr[i]
  end
  map.size.times do |i|
    if map.has_key?(map[arr[i]])
      puts [arr[i], map[arr[i]]]
      return
    end
  end
  puts []
end

def handle_input
  print "Enter an array: "
  arr = gets.chomp!.gsub(/\]|\[/, '').split(',').map!{|ele| ele.to_i}
  print "Enter a number: "
  num = gets.chomp!.to_i
  [arr, num]
end

def main
  puts "** Pair Sum N problem **"
  arr, num = handle_input
  pair_sum(arr, num)
end

---

Question 2 - Max Subarray Sum

JavaScript Implementation

Solution 1 by @MadhavBahl

Method - Bruteforce Search – Worst Case Complexity = O(m.n), where m = array length and n is input number

/**
 * @author MadhavBahl
 * @date 17/01/2019
 * Method - Bruteforce Search
 */

function maxSubarraySum (arr, n) {
    let max = 0;

    for (let i=0; i<arr.length - n + 1; i++) {
        let currentSum = 0;

        for (let j=i; j<i+n; j++) 
            currentSum += arr[j];
        
        if (currentSum > max)
            max = currentSum;
    }

    return max;
}

console.log (maxSubarraySum ([1, 3, 2, 4, 7, 5, 4], 3)); // 16

Solution 2 by @MadhavBahl

Method - Sliding Wndow – Solvable in O(m) time, m = array length

/**
 * @author MadhavBahl
 * @date 17/01/2019
 * Method - Sliding Wndow -- Solvable in O(m) time, m = array length
 */

function maxSubarraySum (arr, n) {
    let currentSum = 0;

    for (let i=0; i<n; i++) 
        currentSum += arr[i];

    let  max = currentSum;

    for (let i=n; i<arr.length; i++) {
        currentSum -= arr[i-n];
        currentSum += arr[i];

        if (currentSum > max)
            max = currentSum;
    }

    return max;
}

console.log (maxSubarraySum ([1, 3, 2, 4, 7, 5, 4], 3)); // 16

JavaScript Implementation

Solution 1 by @MadhavBahl

/**
 * @author MadhavBahl
 * @date 17/01/2019
 * Method used -- bruteforce search (Iteration using nested loops)
 */

function pairSumN (arr, num) {
    let pairArray = [];

    for (let i=0; i<arr.length-1; i++)
        for (let j=i+1; j<arr.length; j++) {
            if (arr[i] + arr[j] === num) {
                pairArray.push (arr[i], arr[j]);
                return pairArray;
            }
        }
    
    return pairArray;
}

console.log (pairSumN ([-3, -2, -1, 0, 1, 2, 3], 0));
console.log (pairSumN ([-1, 4, 5, 5, 6, 7, 8], 10));
console.log (pairSumN ([1, 2, 3, 4, 5, 6], 20));

Ruby Implementation

Solution

=begin
@author: aaditkamat
@date: 17/01/2019
=end
require_relative 'pair_sum_n'

def max_subarray_sum(arr, num)
  #sort array in descending order
  arr.sort!{|x, y| y <=> x}
  print arr.take(num).sum
end
def main
  puts "** Max Subarray problem **"
  arr, num = handle_input
 max_subarray_sum(arr, num)
end

main