diff --git a/Problems/3108 - Minimum Cost Walk in Weighted Graph - Hard/Explanation.md b/Problems/3108 - Minimum Cost Walk in Weighted Graph - Hard/Explanation.md
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+# Comprehensive Explanations of Programs
+
+---
+
+## **Program 1: Longest Nice Subarray**
+
+### **Objective**
+The goal of the program is to find the length of the longest "nice" subarray within the input array `nums`. A subarray is considered "nice" if the bitwise AND operation between every pair of elements in the subarray results in `0`. The program efficiently calculates this using a sliding window approach.
+
+---
+
+### **Step-by-Step Explanation**
+
+#### **1. Initialize Variables**
+int maxLength = 1, left = 0, usedBits = 0;
+
+#### **2. Iterate Over the Array**
+for (int right = 0; right < nums.length; right++) {
+    while ((usedBits & nums[right]) != 0) {
+        usedBits ^= nums[left++];
+    }
+    usedBits |= nums[right];
+    maxLength = Math.max(maxLength, right - left + 1);
+}
+Outer Loop (for): Iterates through each element of the array using the right pointer, which represents the end of the current window.
+
+Inner Loop (while):
+(usedBits & nums[right]) != 0: Checks if adding the current element (nums[right]) to the subarray would result in a bitwise AND operation that is non-zero, which violates the "nice" condition.
+
+usedBits ^= nums[left++]: Removes the element at the left pointer from the subarray by performing a bitwise XOR, effectively shrinking the window from the left until the subarray is "nice" again.
+
+Update Subarray:
+usedBits |= nums[right]: Adds the current element (nums[right]) to the subarray by updating the usedBits variable using a bitwise OR operation.
+
+maxLength = Math.max(maxLength, right - left + 1): Updates the maximum length of the "nice" subarray encountered so far.
+
+Make DSU of size n by using edges vector.After thet we will have each node parent as if nodes are in same component then parent will be same else parent will be different.
+
+After that we will precompute AND for all nodes and store it in ands vector.
+One crucial thing is that we will store AND of component at component's main node(parent) index in ands vector.
+
+Final step: If parent of both query node is same then store AND of that component in ans vector.else store -1.
+
+Complexity
+Time complexity: O(E+Q)
+E : Size of edges vector
+Q : Size of query vector
+
+Space complexity:O(N)
+N : No. of nodes
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diff --git a/Problems/3108 - Minimum Cost Walk in Weighted Graph - Hard/Solution.java b/Problems/3108 - Minimum Cost Walk in Weighted Graph - Hard/Solution.java
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+++ b/Problems/3108 - Minimum Cost Walk in Weighted Graph - Hard/Solution.java	
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+import java.util.*;
+
+class Solution {
+    class DSU {
+        int[] par, rank;
+        
+        public DSU(int n) {
+            par = new int[n + 1];
+            rank = new int[n + 1];
+            Arrays.fill(rank, 1);
+            for (int i = 0; i <= n; i++) par[i] = i;
+        }
+
+        public int find_set(int v) {
+            if (v == par[v]) return v;
+            return par[v] = find_set(par[v]); // Path compression
+        }
+
+        public void union_sets(int a, int b) {
+            a = find_set(a);
+            b = find_set(b);
+            if (a != b) {
+                if (rank[a] < rank[b]) {
+                    int temp = a;
+                    a = b;
+                    b = temp;
+                }
+                par[b] = a;
+                rank[a] += rank[b];
+            }
+        }
+    }
+
+    public List<Integer> minimumCost(int n, int[][] edges, int[][] query) {
+        DSU ds = new DSU(n);
+        int[] ands = new int[n + 1];
+        Arrays.fill(ands, -1);
+
+        for (int[] it : edges) {
+            ds.union_sets(it[0], it[1]);
+        }
+
+        for (int[] it : edges) {
+            int root = ds.find_set(it[0]);
+            int cur = ands[root];
+            ands[root] = (cur == -1) ? it[2] : (cur & it[2]);
+        }
+
+        List<Integer> ans = new ArrayList<>();
+        for (int[] it : query) {
+            if (ds.find_set(it[0]) == ds.find_set(it[1])) {
+                ans.add(ands[ds.find_set(it[0])]);
+            } else {
+                ans.add(-1);
+            }
+        }
+        return ans;
+    }
+}
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