Javascript easy solution with Example
Intuition\nImagine we have a list of numbers like this: [-1, 1, 2, 3, 1], and we also have a target number, let\'s say 2. The goal is to find pairs of numbers f
# Intuition\nImagine we have a list of numbers like this: [-1, 1, 2, 3, 1], and we also have a target number, let\'s say 2. The goal is to find pairs of numbers from the list such that when we add those numbers together, the result is less than the target number.\n\nstep-by-step explanation:\n\n**List of Numbers**: list of numbers is [-1, 1, 2, 3, 1].\n\n**Target Number**: The target number is 2.\n\n**Pairs and Sums:** We will consider each possible pair of numbers and calculate their sum. Then, we will check if that sum is less than the target number.\n\n- For the first pair (0, 1) where the numbers are -1 and 1, the sum is -1 + 1 = 0, which is less than the target number 2.\n- For the second pair (0, 2) where the numbers are -1 and 2, the sum is -1 + 2 = 1, which is less than the target number 2.\n- For the third pair (0, 3) where the numbers are -1 and 3, the sum is -1 + 3 = 2, which is not less than the target number 2.\n- For the fourth pair (0, 4) where the numbers are -1 and 1, the sum is -1 + 1 = 0, which is less than the target number 2.\n- ... and so on for other pairs.\n\n**Counting:** count the pairs where the sum of the numbers is less than the target number. In this example, there are three pairs that satisfy this condition: (0, 1), (0, 2), and (0, 4).\n\n# Approach\nThe problem is about finding pairs of numbers in a list, calculating their sum, and counting how many of these sums are less than a given target number. The solution involves comparing the sum with the target and keeping track of how many times this comparison holds true.\n\n# Complexity\n- Time complexity:O(n^2)\n\n- Space complexity: O(1)\n\n# Code\n```\n/**\n * @param {number[]} nums\n * @param {number} target\n * @return {number}\n */\nvar countPairs = function(nums, target) {\n      let count = 0;\n\n      for (let i = 0; i < nums.length; i++) {\n        for (let j = i + 1; j < nums.length; j++) {\n          if (i < j && nums[i] + nums[j] < target) {\n            count++;\n          }\n        }\n      }\n\n      return count;\n};\n```
['JavaScript']