Solution Leetcode and Explaination Every single line

Author: Github-Saurabh0

61. Container With Most Water/Program Explaination.txt

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+✅ Java Code (Optimal Two-Pointer Approach)
+
+class Solution {
+ public int maxArea(int[] height) {
+ int max = 0;
+ int left = 0, right = height.length - 1;
+
+ while (left < right) {
+ int h = Math.min(height[left], height[right]);
+ int w = right - left;
+ max = Math.max(max, h * w);
+
+ // Move pointer which has smaller height
+ if (height[left] < height[right]) {
+ left++;
+ } else {
+ right--;
+ }
+ }
+
+ return max;
+ }
+}
+
+
+📘 Explanation (Line-by-Line)
+
+int max = 0;
+🔹 Ab tak ka maximum area store karne ke liye variable.
+
+int left = 0, right = height.length - 1;
+🔹 2 pointer technique: Ek array ke start par (left) aur ek end par (right).
+
+while (left < right) {
+🔹 Jab tak left aur right pointers ek dusre ko cross nahi karte.
+
+ int h = Math.min(height[left], height[right]);
+ 🔹 Dono boundaries me se chhoti height lete hain (kyunki water wahin tak bharega).
+
+ int w = right - left;
+ 🔹 Width nikaali → right aur left ke beech ka distance.
+
+ max = Math.max(max, h * w);
+ 🔹 Max area update kiya (height × width).
+
+ if (height[left] < height[right]) {
+ left++;
+ 🔹 Chhoti boundary ko move karo (chance zyada area ka badhne ka hai).
+ } else {
+ right--;
+ 🔹 Right side chhoti hai toh usse move karo.
+ }
+}
+
+return max;
+🔹 Sabse bada area return kar do.
+
+🧠 Example Input:
+Input: height = [1,8,6,2,5,4,8,3,7]
+Output: 49
+Explanation: Water is stored between index 1 and 8 → min(8, 7) × (8 - 1) = 49
+🚀 Time & Space Complexity:
+
+Time: O(n)
+
+Space: O(1)
+
+🔗 Need more help or Java tricks?
+https://www.linkedin.com/in/saurabh884095/

61. Container With Most Water/Solution.java

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+class Solution {
+ public int maxArea(int[] height) {
+ int max = 0;
+ int left = 0, right = height.length - 1;
+
+ while (left < right) {
+ int h = Math.min(height[left], height[right]);
+ int w = right - left;
+ max = Math.max(max, h * w);
+
+ // Move pointer which has smaller height
+ if (height[left] < height[right]) {
+ left++;
+ } else {
+ right--;
+ }
+ }
+
+ return max;
+ }
+}