Added Interleaving Strings Algo

Dynamic Programming/Interleaving Strings/README.md

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+# Interleaving Strings
+
+## Problem Description
+Given three strings `A`, `B`, and `C`, determine if `C` is an interleaving of `A` and `B`. `C` is considered an interleaving of `A` and `B` if it contains all characters of `A` and `B`, with their order preserved, though characters from each string can be interwoven.
+
+### Example
+#### Example 1
+- **Input:**
+A = "aab" B = "axy" C = "aaxaby"
+- **Output:**
+True
+
+- **Explanation:** 
+`C` can be formed by interleaving `A` and `B`, as `C = "aaxaby"` follows the sequence of characters from both `A` and `B`.
+
+#### Example 2
+- **Input:**
+A = "abc" B = "def" C = "abdecf"
+- **Output:**
+False
+- **Explanation:** 
+`C` is not an interleaving of `A` and `B`, as the sequence of characters does not match both `A` and `B`.
+
+## Approach
+We use **Dynamic Programming (DP)** to solve this problem efficiently. We define a DP table where each cell `dp[i][j]` represents whether the first `i` characters of `A` and the first `j` characters of `B` can form the first `i + j` characters of `C`.
+
+### Key Insights
+1. If `A[i-1]` matches `C[i+j-1]`, then `dp[i][j]` depends on `dp[i-1][j]`.
+2. If `B[j-1]` matches `C[i+j-1]`, then `dp[i][j]` depends on `dp[i][j-1]`.
+3. If either condition above holds, then `dp[i][j]` is `true`.
+
+### DP Recurrence Relation
+- If `A[i-1] == C[i+j-1]`, then `dp[i][j] = dp[i-1][j]`.
+- If `B[j-1] == C[i+j-1]`, then `dp[i][j] = dp[i][j-1]`.
+- If either condition is true, `dp[i][j] = true`.
+
+### Steps
+1. Initialize a 2D DP table with dimensions `(n+1) x (m+1)`, where `n` is the length of `A` and `m` is the length of `B`.
+2. Fill the table based on the recurrence relations derived.
+3. The result will be in `dp[n][m]`, representing if all characters of `A` and `B` can interleave to form `C`.
+
+### Complexity Analysis
+- **Time Complexity**: `O(n * m)`, where `n` is the length of `A` and `m` is the length of `B`.
+- **Space Complexity**: `O(n * m)`, for storing the DP table.

Dynamic Programming/Interleaving Strings/program.c

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+#include <stdio.h>
+#include <stdbool.h>
+#include <string.h>
+
+bool isInterleave(char* A, char* B, char* C) {
+    int n = strlen(A);
+    int m = strlen(B);
+    int lenC = strlen(C);
+
+    if (n + m != lenC) {
+        return false;
+    }
+
+    bool dp[n + 1][m + 1];
+
+    dp[0][0] = true;  // Empty A and B to form empty C is true
+
+    for (int j = 1; j <= m; j++) {
+        dp[0][j] = dp[0][j - 1] && (B[j - 1] == C[j - 1]);
+    }
+
+    for (int i = 1; i <= n; i++) {
+        dp[i][0] = dp[i - 1][0] && (A[i - 1] == C[i - 1]);
+    }
+
+    for (int i = 1; i <= n; i++) {
+        for (int j = 1; j <= m; j++) {
+            dp[i][j] = (dp[i - 1][j] && A[i - 1] == C[i + j - 1]) ||
+                       (dp[i][j - 1] && B[j - 1] == C[i + j - 1]);
+        }
+    }
+
+    return dp[n][m];
+}
+
+int main() {
+    char A[] = "aab";
+    char B[] = "axy";
+    char C[] = "aaxaby";
+
+    if (isInterleave(A, B, C)) {
+        printf("True\n");
+    } else {
+        printf("False\n");
+    }
+
+    return 0;
+}