diff --git a/Daily Questions/#2379 - Minimum Recolors to Get K Consecutive Black Blocks - Easy/Exaplanation.md b/Daily Questions/#2379 - Minimum Recolors to Get K Consecutive Black Blocks - Easy/Exaplanation.md
new file mode 100644
index 0000000..7a29325
--- /dev/null
+++ b/Daily Questions/#2379 - Minimum Recolors to Get K Consecutive Black Blocks - Easy/Exaplanation.md	
@@ -0,0 +1,19 @@
+n: The length of the input string blocks.
+
+b: A counter to keep track of the number of black blocks within a sliding window.
+
+Max: To store the maximum number of black blocks found in any sliding window of length k.
+
+This loop counts the number of black blocks ('B') in the first window of length k and assigns it to both b and Max.
+
+This loop slides the window across the string from index 0 to n - k - 1.
+
+It decreases the count b if the block at the start of the window is 'B'.
+
+It increases the count b if the block at the end of the window is 'B'.
+
+Updates Max to the maximum value between the current Max and b.
+
+If Max becomes equal to or greater than k, it means we already have k consecutive black blocks, and hence, no recoloring is needed. The function returns 0.
+
+If the loop completes without finding k consecutive black blocks, the function returns the difference between k and Max, which represents the minimum number of recolors needed.
\ No newline at end of file
diff --git a/Daily Questions/#2379 - Minimum Recolors to Get K Consecutive Black Blocks - Easy/Solution.java b/Daily Questions/#2379 - Minimum Recolors to Get K Consecutive Black Blocks - Easy/Solution.java
new file mode 100644
index 0000000..f7758cb
--- /dev/null
+++ b/Daily Questions/#2379 - Minimum Recolors to Get K Consecutive Black Blocks - Easy/Solution.java	
@@ -0,0 +1,30 @@
+class Solution {
+    public int minimumRecolors(String blocks, int k) {
+        int n = blocks.length(), b = 0, Max;
+
+        for(int i = 0; i < k; i++){
+            if(blocks.charAt(i) == 'B'){
+                b++;
+            }
+        }
+
+        Max = b;
+
+        for(int i = 0; i < n - k; i++){
+            if(blocks.charAt(i) == 'B'){
+                b--;
+            }
+            if(blocks.charAt(i + k) == 'B'){
+                b++;
+            }
+
+            Max = Math.max(Max, b);
+
+            if(Max >= k){
+                return 0;
+            }
+        }
+
+        return k - Max;
+    }
+}
\ No newline at end of file
diff --git a/Daily Questions/#2523 - Closest Prime Numbers in Range - Medium/Explanation.md b/Daily Questions/#2523 - Closest Prime Numbers in Range - Medium/Explanation.md
new file mode 100644
index 0000000..ef2180b
--- /dev/null
+++ b/Daily Questions/#2523 - Closest Prime Numbers in Range - Medium/Explanation.md	
@@ -0,0 +1,34 @@
+private static boolean[] isPrime = new boolean[1000001]; This boolean array is used to mark prime numbers up to 1000000. The size of the array is 1000001 to include the number 1000000.
+
+Static Block:
+The static block is used to initialize the isPrime array and precompute the prime numbers.
+
+Arrays.fill(isPrime, true); This method sets all elements of the isPrime array to true, assuming all numbers are prime initially.
+
+precompute(); This method is called to mark non-prime numbers in the isPrime array.
+
+Method: precompute()
+This method uses the Sieve of Eratosthenes algorithm to mark non-prime numbers in the isPrime array.
+
+isPrime[0] = isPrime[1] = false; Numbers 0 and 1 are not prime.
+
+A loop is used to iterate from 2 to sqrt(1000000).
+
+If isPrime[i] is true, all multiples of i are marked as false (not prime) starting from i*i.
+
+Method: closestPrimes(int left, int right)
+This method finds the closest prime numbers within the range [left, right].
+
+int prev = -1, minDiff = Integer.MAX_VALUE; Variables to track the previous prime number and the minimum difference between consecutive primes.
+
+int num1 = -1, num2 = -1; Variables to store the closest prime numbers.
+
+A loop iterates through the range [left, right].
+
+If isPrime[i] is true, the number i is a prime.
+
+If prev is not -1 and the difference (i - prev) is less than minDiff, update minDiff and set num1 to prev and num2 to i.
+
+Update prev to i.
+
+If no prime numbers are found, return new int[]{-1, -1}. Otherwise, return the closest prime numbers num1 and num2.
\ No newline at end of file
diff --git a/Daily Questions/#2523 - Closest Prime Numbers in Range - Medium/Solution.java b/Daily Questions/#2523 - Closest Prime Numbers in Range - Medium/Solution.java
new file mode 100644
index 0000000..2e417a5
--- /dev/null
+++ b/Daily Questions/#2523 - Closest Prime Numbers in Range - Medium/Solution.java	
@@ -0,0 +1,39 @@
+import java.util.*;
+
+class Solution {
+    private static boolean[] isPrime = new boolean[1000001];
+
+    static {
+        Arrays.fill(isPrime, true);
+        precompute();
+    }
+
+    private static void precompute() {
+        isPrime[0] = isPrime[1] = false;
+        for (int i = 2; i * i <= 1000000; i++) {
+            if (isPrime[i]) {
+                for (int j = i * i; j <= 1000000; j += i) {
+                    isPrime[j] = false;
+                }
+            }
+        }
+    }
+
+    public int[] closestPrimes(int left, int right) {
+        int prev = -1, minDiff = Integer.MAX_VALUE;
+        int num1 = -1, num2 = -1;
+
+        for (int i = left; i <= right; i++) {
+            if (isPrime[i]) {
+                if (prev != -1 && (i - prev < minDiff)) {
+                    minDiff = i - prev;
+                    num1 = prev;
+                    num2 = i;
+                }
+                prev = i;
+            }
+        }
+
+        return (num1 == -1) ? new int[]{-1, -1} : new int[]{num1, num2};
+    }
+}
\ No newline at end of file