diff --git a/algos/range_queries/PrefixSum/soln/Advait/Solution1.cpp b/algos/range_queries/PrefixSum/soln/Advait/Solution1.cpp
new file mode 100644
index 0000000..6aa297d
--- /dev/null
+++ b/algos/range_queries/PrefixSum/soln/Advait/Solution1.cpp
@@ -0,0 +1,114 @@
+/*
+------------------------------------------------------------
+LeetCode Problem: 209 - Minimum Size Subarray Sum
+Difficulty: Medium
+Category: Prefix Sum / Sliding Window
+------------------------------------------------------------
+
+Problem Statement:
+Given an array of positive integers nums and an integer target,
+return the minimal length of a contiguous subarray of which the
+sum is greater than or equal to target. If there is no such
+subarray, return 0.
+
+------------------------------------------------------------
+Approach 1: Prefix Sum + Two Pointers (Explicit Prefix Array)
+------------------------------------------------------------
+
+Idea:
+- Precompute prefix sums to allow O(1) subarray sum calculation.
+- Use two pointers on the prefix array to maintain a sliding window.
+- Since all elements are positive, expanding the right pointer
+  increases the sum and moving the left pointer decreases it.
+
+Subarray Sum Formula:
+sum(j..i-1) = prefix[i] - prefix[j]
+
+Time Complexity: O(N)
+Space Complexity: O(N)
+
+------------------------------------------------------------
+*/
+
+#include <bits/stdc++.h> 
+using namespace std; 
+class Solution { 
+    public: 
+    int minSubArrayLen(int target, vector<int>& nums) { 
+        vector<int> a (nums.size()+1); 
+        a[0]=0; 
+        for(int i=0;i<nums.size();i++){ 
+            a[i+1]=a[i]+nums[i]; //prefix sum
+        } 
+        int i=1,j=0; 
+        int minsize=nums.size(); 
+        while(i<nums.size()+1){ 
+            int currsum=a[i]-a[j]; 
+            if(currsum<target){ 
+                i++; //update pointer if sum less than target
+            }
+            else{ 
+                if(minsize>(i-j)){
+                    minsize=i-j; 
+                } 
+                j++; //increase lower pointer to find more possible subarrays
+            } 
+        } 
+        if(a[nums.size()]>=target){ 
+            return minsize; //if total cumulative sum is less than target then target achieving is not possible
+        } 
+        else { 
+            return 0; 
+        } 
+    } 
+};
+/*
+------------------------------------------------------------
+Optimized Approach: Sliding Window (Implicit Prefix Sum)
+------------------------------------------------------------
+
+Idea:
+- Instead of storing the entire prefix sum array, maintain
+  the current subarray sum directly.
+- This is equivalent to prefix sum logic but reduces
+  extra space usage.
+
+Key Insight:
+- Because all elements are positive, the window can safely
+  move forward without missing optimal solutions.
+
+Time Complexity: O(N)
+Space Complexity: O(1)
+
+------------------------------------------------------------
+*/
+
+class Solution {
+public:
+    int minSubArrayLen(int target, vector<int>& nums) {
+        int sum=nums[0];
+        int minsize=nums.size();
+        bool flag=false;//set flag for if target is not achievable
+        int i=0,j=0;
+        while(i<nums.size()){
+            if(sum<target){
+                i++;
+                if(i<nums.size()){
+                    sum+=nums[i];//increase sum
+                }
+            }
+            else{
+                flag=true;
+                if(minsize>(i-j+1)){
+                    minsize=i-j+1;//update minsize if its lesser
+                }
+                sum-=nums[j];
+                j++;//decrease window
+            }
+        }
+        if(flag){
+            return minsize;
+        }
+        return 0;
+    }
+};
diff --git a/algos/range_queries/PrefixSum/soln/Advait/Solution2.cpp b/algos/range_queries/PrefixSum/soln/Advait/Solution2.cpp
new file mode 100644
index 0000000..e6e64ca
--- /dev/null
+++ b/algos/range_queries/PrefixSum/soln/Advait/Solution2.cpp
@@ -0,0 +1,133 @@
+/*
+------------------------------------------------------------
+LeetCode Problem: 560 - Subarray Sum Equals K
+Difficulty: Medium
+Category: Prefix Sum
+------------------------------------------------------------
+
+Problem Statement:
+Given an integer array nums and an integer k, return the total
+number of continuous subarrays whose sum equals k.
+
+------------------------------------------------------------
+Approach: Prefix Sum + Brute Force Enumeration
+------------------------------------------------------------
+
+Idea:
+- Compute the prefix sum array where prefix[i] stores the sum of
+  elements from index 0 to i-1.
+- Any subarray sum from index i to j-1 can be computed in O(1) as:
+  
+        prefix[j] - prefix[i]
+
+- Enumerate all possible (i, j) pairs and count how many satisfy
+  prefix[j] - prefix[i] == k.
+
+Why this works:
+- Prefix sums allow constant-time subarray sum calculation.
+- Checking all subarrays guarantees correctness even with
+  negative numbers and zeros.
+
+------------------------------------------------------------
+Time Complexity:
+- Prefix sum construction: O(N)
+- Nested loops over all subarrays: O(N²)
+- Total: O(N²)
+
+Space Complexity:
+- Prefix sum array: O(N)
+
+------------------------------------------------------------
+*/
+#include <bits/stdc++.h>
+using namespace std;
+
+class Solution {
+public:
+    int subarraySum(vector<int>& nums, int k) {
+        vector<int> a (nums.size()+1); 
+        a[0]=0; 
+        for(int i=0;i<nums.size();i++){ 
+            a[i+1]=a[i]+nums[i];
+        } 
+        int count=0;
+        for(int i=0;i<nums.size()+1;i++){
+            for(int j=i+1;j<nums.size()+1;j++){
+                if(a[j]-a[i]==k){
+                    count++;
+                }
+            }
+        }
+        return count;
+    }
+};
+
+/*
+------------------------------------------------------------
+LeetCode Problem: 560 - Subarray Sum Equals K
+Difficulty: Medium
+Category: Prefix Sum + HashMap
+------------------------------------------------------------
+
+Problem Statement:
+Given an integer array nums and an integer k, return the total
+number of continuous subarrays whose sum equals k.
+
+------------------------------------------------------------
+Approach: Prefix Sum with Frequency HashMap
+------------------------------------------------------------
+
+Key Observation:
+For a subarray ending at index j with sum equal to k:
+
+    prefix[j] - prefix[i] = k
+=>  prefix[i] = prefix[j] - k
+
+Thus, for each prefix sum encountered, we need to know how many
+times (prefixSum - k) has appeared previously.
+
+Algorithm:
+1. Traverse the array once while maintaining a running prefix sum.
+2. Use a hashmap to store the frequency of prefix sums seen so far.
+3. At each index:
+   - Check if (currentPrefixSum - k) exists in the hashmap.
+   - If it exists, its frequency contributes to the answer.
+4. Update the frequency of the current prefix sum.
+
+Important Detail:
+- Initialize the hashmap with {0 : 1} to handle subarrays that
+  start from index 0.
+
+------------------------------------------------------------
+Correctness:
+- This method counts all valid subarrays ending at each index.
+- Works correctly even with negative numbers and zeros.
+- Avoids the need for nested loops by reusing prefix sum history.
+
+------------------------------------------------------------
+Time Complexity:
+- O(N), where N is the size of the array.
+
+Space Complexity:
+- O(N) for storing prefix sum frequencies in the hashmap.
+
+------------------------------------------------------------
+*/
+
+class Solution {
+public:
+    int subarraySum(vector<int>& nums, int k) {
+        unordered_map <int,int> fre;
+        fre[0]=1;
+        int sum=0;
+        int count=0;
+        for(int x=0;x<nums.size();x++){
+            sum+=nums[x];
+            if(fre.find(sum-k)!=fre.end()){
+                count+=fre[sum-k];
+            }
+            fre[sum]++;
+        }
+        return count;
+    }
+};