Solution - #167 - Hans - 07/04/2025

Arrays & Strings/Two Sum/Explanation.md

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+Explanation (C solution):
+
+Two Sum II, a slightly modified version of Two Sum I where we get a sorted array. 
+The solution's straightforward with the utilization of two pointers - at the start and end of the array respectively, let's say l and r.
+We move inwards into the array until our start and end pointers don't cross over i.e left > right.
+We check if the value at array[l] + array[r] (which is our sum) == target.
+
+Since it's a sorted array. If: 
+1. Sum is greater than Target
+-Then we know we need a smaller value to match or get close to the target, hence we decrease the end pointer , pointing to a smaller value (second largest value and so on..).
+2. Sum is lesser than Target
+-Then we need a larger value to match or get close to the target, hence we increase the start pointer, pointing to a larger value(second smallest value and so on..).
+
+If Sum is equal to our target:
+-Store the indexes of the two values in the malloced array (dynamic array since we can't directly return two values from a function in C)
+-We've increased the index by 1 to facilitate 1-based indexing as given in the problem.
+-Return the malloced array.
+
+Time Complexity: O(n)

Arrays & Strings/Two Sum/Two Sum 2.c

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+
+int* twoSum(int* numbers, int numbersSize, int target, int* returnSize) {
+    int l = 0; 
+    int r = numbersSize-1;
+
+    int* answer = (int*)malloc(2*sizeof(int)); //dynamic memory allocation
+    *returnSize = 2; //we're returning two values
+
+    while (l < r)
+    {
+        int sum = (numbers[l] + numbers[r]);
+        if (sum == target)
+        {
+            answer[0] = l+1; //facilitating 1-based indexing as required by the problem.
+            answer[1] = r+1;
+            return answer;
+        }
+        else if (sum > target)
+        {
+            r--; //point to a smaller value to reduce the sum
+        }
+        else
+        {
+            l++; //point to a larger value to increase the sum
+        }
+
+    }
+    return 0;
+
+}