In this course, we will cover coding problems that utilize the two-pointer algorithm and explain the process of solving the problems step by step. Given an integer array To solve this problem, we will approach it by using the two-pointer technique to traverse the array while Initialize the starting pointer For example, use the following loop to compare the sum of the values pointed to by the two pointers:<\/p>\n Once we find valid indices for
\n The two-pointer technique involves using two pointers in linear data structures such as arrays or lists, reducing time complexity and enabling
\n efficient problem solving.<\/p>\n<\/header>\nProblem Description<\/h2>\n
nums<\/code> and an integer target<\/code>,
\n solve the problem of finding the indices of two numbers in the nums<\/code> array that add up to equal target<\/code> and return those indices.
\n Users can assume that there is exactly one pair of integers that exists. <\/p>\nInput Example<\/h3>\n
\n
nums = [2, 7, 11, 15]<\/code><\/li>\ntarget = 9<\/code><\/li>\n<\/ul>\nOutput Example<\/h3>\n
[0, 1]<\/code><\/p>\n<\/section>\nAlgorithm Approach<\/h2>\n
\n comparing the sum of the values pointed to by the two pointers to find the desired value.
\n The two-pointer technique is typically applied to sorted arrays, but in this problem, the two pointers
\n can point to the same array while considering various cases.<\/p>\n<\/section>\nProblem Solving Process<\/h2>\n
1. Set initial indices for the two pointers in the array<\/h3>\n
left<\/code> to 0 and the ending pointer right<\/code> to the length of the array – 1.
\n This allows us to explore the array through nums[left]<\/code> and nums[right]<\/code>.<\/p>\n2. Move pointers in a loop<\/h3>\n
while (left < right) {\n int sum = nums[left] + nums[right];\n if (sum == target) {\n return new int[] { left, right };\n } else if (sum < target) {\n left++;\n } else {\n right--;\n }\n }<\/code><\/pre>\n3. Return the result<\/h3>\n
left<\/code> and right<\/code>, we return those values.
\n If no valid values are found, handle it by returning null<\/code> or throwing an exception.<\/p>\n<\/section>\nC# Code Implementation<\/h2>\n
\nusing System;\n\nclass Program {\n static void Main(string[] args) {\n int[] nums = { 2, 7, 11, 15 };\n int target = 9;\n int[] result = TwoSum(nums, target);\n Console.WriteLine($\"[{result[0]}, {result[1]}]\");\n }\n\n static int[] TwoSum(int[] nums, int target) {\n int left = 0;\n int right = nums.Length - 1;\n\n while (left < right) {\n int sum = nums[left] + nums[right];\n if (sum == target) {\n return new int[] { left, right };\n } else if (sum < target) {\n left++;\n } else {\n right--;\n }\n }\n\n \/\/ If no matching pair is found\n throw new Exception(\"No two sum solution\");\n }\n}\n <\/code><\/pre>\n<\/section>\nSolution Analysis<\/h2>\n
1. Time Complexity<\/h3>\n