Frog jump

src/climbing-stairs/solutions/memoization.ts

@@ -1,18 +1,13 @@
 import { MemoObjectType } from '../../@types/types';
 
-/**
-
-
- */
-
 /**
  *
  * @param {number} n
  * @returns {number}
  */
 
 // Solution 1
-const solution = (n: number): number => {
+const climbingStairs1 = (n: number): number => {
   let m: MemoObjectType = {};
   const backtrack = (n: number) => {
     if (n <= 1) return 1;
@@ -35,12 +30,12 @@ const solution = (n: number): number => {
 // Solution 2 (Having a default hash,
 // no need to declare it in scope of the function)
 
-const solution2 = (n: number, m: MemoObjectType = {}): number => {
+const climbingStairs2 = (n: number, m: MemoObjectType = {}): number => {
   if (n <= 1) return 1;
 
   if (m[n]) return m[n];
 
-  m[n] = solution2(n - 1, m) + solution2(n - 2, m);
+  m[n] = climbingStairs2(n - 1, m) + climbingStairs2(n - 2, m);
   return m[n];
 };
 
@@ -52,12 +47,15 @@ const solution2 = (n: number, m: MemoObjectType = {}): number => {
  */
 
 // Solution 3 (Having a default array, no need to declare it in scope of the function)
-const solution3 = (n: number, m: number[] = Array(n + 1).fill(-1)): number => {
+const climbingStairs3 = (
+  n: number,
+  m: number[] = Array(n + 1).fill(-1)
+): number => {
   if (n <= 1) return 1;
 
   if (m[n] != -1) return m[n];
 
-  m[n] = solution3(n - 1, m) + solution3(n - 2, m);
+  m[n] = climbingStairs3(n - 1, m) + climbingStairs3(n - 2, m);
   return m[n];
 };
 
@@ -68,7 +66,7 @@ const solution3 = (n: number, m: number[] = Array(n + 1).fill(-1)): number => {
  */
 
 // Solution 4 (Using array for memoization)
-const solution4 = (n: number): number => {
+const climbingStairs4 = (n: number): number => {
   let m = Array(n + 1).fill(-1);
   const backtrack = (n: number) => {
     if (n <= 1) return 1;

src/climbing-stairs/solutions/tabulation.ts

@@ -12,7 +12,7 @@
  * of memoization)
  */
 
-const solutionOne = (n: number): number => {
+const climbingStairs1 = (n: number): number => {
   function* fibGenerator() {
     let arr = [1, 1];
 
@@ -43,7 +43,7 @@ const solutionOne = (n: number): number => {
  * @returns {number}
  */
 
-const solutionTwo = (n: number): number => {
+const climbingStairs2 = (n: number): number => {
   let dp = [1, 1];
 
   for (let i = 2; i <= n; i++) {
@@ -65,7 +65,7 @@ const solutionTwo = (n: number): number => {
  * @returns {number}
  */
 
-const soultionThree = (n: number): number => {
+const climbingStairs3 = (n: number): number => {
   let prev1 = 1;
   let prev2 = 1;
 

src/fibonacci/docs/PROBLEM.md

@@ -12,5 +12,5 @@ Given n, calculate F(n).
 
 ## Solutions
 
-* Tabulation: A ‘bottom-up’ approach where we start from the base case and reach the final answer that we want.
-* Memoization: A top-down approach where we cache the results of function calls and return the cached result if the function is called again with the same inputs.
+- Tabulation: A ‘bottom-up’ approach where we start from the base case and reach the final answer that we want.
+- Memoization: A top-down approach where we cache the results of function calls and return the cached result if the function is called again with the same inputs and the same result.

src/fibonacci/solutions/memoization.ts

@@ -7,7 +7,7 @@ import { MemoObjectType } from '../../@types/types';
  */
 
 // Solution 1
-const solution = (n: number): number => {
+const fib1 = (n: number): number => {
   let m: MemoObjectType = {};
   const backtrack = (n: number) => {
     if (n <= 1) return n;
@@ -30,12 +30,12 @@ const solution = (n: number): number => {
 // Solution 2 (Having a default hash,
 // no need to declare it in scope of the function)
 
-const solution2 = (n: number, m: MemoObjectType = {}): number => {
+const fib2 = (n: number, m: MemoObjectType = {}): number => {
   if (n <= 1) return n;
 
   if (m[n]) return m[n];
 
-  m[n] = solution2(n - 1, m) + solution2(n - 2, m);
+  m[n] = fib2(n - 1, m) + fib2(n - 2, m);
   return m[n];
 };
 
@@ -47,12 +47,12 @@ const solution2 = (n: number, m: MemoObjectType = {}): number => {
  */
 
 // Solution 3 (Having a default array, no need to declare it in scope of the function)
-const solution3 = (n: number, m: number[] = Array(n + 1).fill(-1)): number => {
+const fib3 = (n: number, m: number[] = Array(n + 1).fill(-1)): number => {
   if (n <= 1) return n;
 
   if (m[n] != -1) return m[n];
 
-  m[n] = solution3(n - 1, m) + solution3(n - 2, m);
+  m[n] = fib3(n - 1, m) + fib3(n - 2, m);
   return m[n];
 };
 
@@ -63,7 +63,7 @@ const solution3 = (n: number, m: number[] = Array(n + 1).fill(-1)): number => {
  */
 
 // Solution 4 (Using array for memoization)
-const solution4 = (n: number): number => {
+const fib4 = (n: number): number => {
   let m = Array(n + 1).fill(-1);
   const backtrack = (n: number) => {
     if (n <= 1) return n;

src/fibonacci/solutions/tabulation.ts

@@ -12,7 +12,7 @@
  * of memoization)
  */
 
-const solution1 = (n: number): number => {
+const fib1 = (n: number): number => {
   function* fibGenerator() {
     let arr = [0, 1];
 
@@ -43,7 +43,7 @@ const solution1 = (n: number): number => {
  * @returns {number}
  */
 
-const solution2 = (n: number): number => {
+const fib2 = (n: number): number => {
   let dp = [0, 1];
 
   for (let i = 2; i <= n; i++) {
@@ -65,7 +65,7 @@ const solution2 = (n: number): number => {
  * @returns {number}
  */
 
-const soultion3 = (n: number): number => {
+const fib3 = (n: number): number => {
   let prev1 = 0;
   let prev2 = 1;
 

src/frog-jump/docs/PROBLEM.md

@@ -0,0 +1,10 @@
+# Frog Jump 🐸
+
+There is a frog on the '1st' step of an 'N' stairs long staircase. The frog wants to reach the 'Nth' stair. 'HEIGHT[i]' is the height of the '(i+1)th' stair.If Frog jumps from 'ith' to 'jth' stair, the energy lost in the jump is given by absolute value of ( HEIGHT[i-1] - HEIGHT[j-1] ). If the Frog is on 'ith' staircase, he can jump either to '(i+1)th' stair or to '(i+2)th' stair. Your task is to find the minimum total energy used by the frog to reach from '1st' stair to 'Nth' stair.
+
+[CodeNinja](https://www.codingninjas.com/studio/problems/frog-jump_3621012)
+
+## Solutions
+
+- Tabulation: A ‘bottom-up’ approach where we start from the base case and reach the final answer that we want.
+- Memoization: A top-down approach where we cache the results of function calls and return the cached result if the function is called again with the same inputs and the same result.

src/frog-jump/solutions/memoization.ts

@@ -0,0 +1,55 @@
+/**
+ *
+ * @param {number} n
+ * @param {number[]} heights
+ * @returns {number}
+ */
+
+const fragJump1 = (n: number, heights: number[]): number => {
+  if (n === 0) {
+    return 0;
+  }
+
+  let jumpOne =
+    fragJump1(n - 1, heights) + Math.abs(heights[n] - heights[n - 1]);
+  let jumpTwo = Infinity;
+
+  if (n > 1) {
+    jumpTwo = fragJump1(n - 2, heights) + Math.abs(heights[n] - heights[n - 2]);
+  }
+
+  return Math.min(jumpOne, jumpTwo);
+};
+
+/**
+ * An Optimized Solution of the above.
+ *
+ * @param {number} n
+ * @param {number[]} heights
+ * @param {number[]} dp
+ * @returns {number}
+ */
+
+const fragJump2 = (
+  n: number,
+  heights: number[],
+  dp: number[] = Array(n + 1).fill(-1)
+): number => {
+  if (n === 0) {
+    return 0;
+  }
+
+  if (dp[n] !== -1) return dp[n];
+
+  let jumpOne =
+    fragJump2(n - 1, heights, dp) + Math.abs(heights[n] - heights[n - 1]);
+  let jumpTwo = Infinity;
+
+  if (n > 1) {
+    jumpTwo =
+      fragJump2(n - 2, heights, dp) + Math.abs(heights[n] - heights[n - 2]);
+  }
+
+  dp[n] = Math.min(jumpOne, jumpTwo);
+  return dp[n];
+};