Top Practice Problem Of Array In Js

  function secondLargest(arr) {
  if (arr.length < 2) return null; // If there are less than 2 elements, return null

  let first = arr[0];  // Assume first element is the largest
  let second = -Infinity; // Initially set second largest as very small

  for (let i = 1; i < arr.length; i++) {
    if (arr[i] > first) {
      second = first; // Move the current largest to second
      first = arr[i]; // Update the largest
    } else if (arr[i] > second && arr[i] !== first) {
      second = arr[i]; // Update second if it's smaller than first but greater than previous second
    }
  }

  return second === -Infinity ? null : second; // If no second largest, return null
}

console.log(secondLargest([10, 5, 20, 8])); // Output: 10
console.log(secondLargest([3, 3, 3, 3])); // Output: null
console.log(secondLargest([5])); // Output: null

  function removeDuplicates(arr) {
  return [...new Set(arr)];
}

console.log(removeDuplicates([1, 2, 2, 3, 4, 4, 5])); 
// Output: [1, 2, 3, 4, 5]

  function arrayIntersection(arr1, arr2) {
  return arr1.filter(value => arr2.includes(value));
}

console.log(arrayIntersection([1, 2, 3, 4], [3, 4, 5, 6])); // Output: [3, 4]
console.log(arrayIntersection(["apple", "banana"], ["banana", "grape"])); // Output: ["banana"]

  function findMax(arr) {
  if (arr.length === 0) return undefined; // Handle empty array case

  let max = arr[0]; // Initialize max with the first element

  for (let i = 1; i < arr.length; i++) { // Start loop from the second element
    if (arr[i] > max) {
      max = arr[i]; // Update max if a larger number is found
    }
  }

  return max;
}

console.log(findMax([10, 5, 20, 8, 30])); // Output: 30
console.log(findMax([-3, -1, -7, -2])); // Output: -1
console.log(findMax([])); // Output: undefined

  function sumArray(arr) {
  let sum = 0;
  for (let i = 0; i < arr.length; i++) {
    sum += arr[i]; // Adding each element to sum
  }
  return sum;
}

console.log(sumArray([1, 2, 3, 4, 5])); // Output: 15