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My Note:-
If you closely observe the pattern for the brute force
const wrapper = (value) => { const helper = (combinedArray, depth) => { if (depth == 3) { // operation return ; } for (let coin of coins) { if (value - coin >=0) { combinedArray.push(coin); helper(combinedArray, label+1); combinedArray.pop(); } } } helper([], 0); return result; }; const res = wrapper(value); console.log(res);
Q1. Start with 2 coin combinations
const wrapper = () => { const coinSide = ['head', 'tail'] const result = []; const helper = (currentCombination, depth) => { if (depth == 2) { result.push([...currentCombination]); return ; } for (side of coinSide) { currentCombination.push(side); helper(currentCombination, depth +1); currentCombination.pop() } } helper([], 0); return result; }; const res = wrapper(); console.log(res);
Q2. Start with 3 coin combinations
const wrapper = () => { const coinSide = ['head', 'tail'] const result = []; const helper = (currentCombination, depth) => { if (depth == 3) { result.push([...currentCombination]); return ; } for (side of coinSide) { currentCombination.push(side); helper(currentCombination, depth +1); currentCombination.pop() } } helper([], 0); return result; }; const res = wrapper(); console.log(res); /* [ [ 'head', 'head', 'head' ], [ 'head', 'head', 'tail' ], [ 'head', 'tail', 'head' ], [ 'head', 'tail', 'tail' ], [ 'tail', 'head', 'head' ], [ 'tail', 'head', 'tail' ], [ 'tail', 'tail', 'head' ], [ 'tail', 'tail', 'tail' ] ] */
Q3. Seating arrangement
const wrapper = () => { const result = []; const group = ['b1', 'b2', 'g1'] const helper = (combination, depth) => { if (depth == 3) { result.push([...combination]); return; } for (let item of group) { if (combination.indexOf(item) < 0) { combination.push(item); helper(combination, depth +1); combination.pop(); } } } helper([], 0); return result; }; /* [ [ 'b1', 'b2', 'g1' ], [ 'b1', 'g1', 'b2' ], [ 'b2', 'b1', 'g1' ], [ 'b2', 'g1', 'b1' ], [ 'g1', 'b1', 'b2' ], [ 'g1', 'b2', 'b1' ] ] */
Q4. Coin / Sum problem
// Minimum coin Problem const wrapper = (value) => { let result = 99999; let resultArr = []; const coins = [10, 6, 1]; const helper = (value, label, combinedArray) => { if (value == 0) { if (result > label) { result = label; resultArr = [...combinedArray] } return ; } for (let coin of coins) { if (value - coin >=0) { combinedArray.push(coin); helper(value-coin, label+1, combinedArray); combinedArray.pop(); } } } helper(value, 0, []); console.log(resultArr) return result; }; const res = wrapper(12); console.log(res); /* [ 6, 6 ] 2 */
Q5.Set generation
// Problem 1: Generating All Subsets of a Set // Problem Statement: // Given a set of unique elements, generate all possible subsets (the power set). // This solution need more enhancement. // Example: // Input: [1, 2, 3] // Output: [[], [1], [2], [3], [1, 2], [1, 3], [2, 3], [1, 2, 3]] const wrapper = () => { const result = [[]]; const input = [1,2,3]; input.forEach(item => result.push([item])); const helper = (combination, depth) => { if (depth == 2) { if (result.indexOf(combination) < 0) { result.push([...combination]); } return; } for (let item of input) { if (combination.indexOf(item) < 0) { combination.push(item); helper(combination, depth+1); combination.pop() } } } helper([], 0); result.push([...input]) return result; } const test = wrapper(); console.log(test); /* [ [], [ 1 ], [ 2 ], [ 3 ], [ 1, 2 ], [ 1, 3 ], [ 2, 1 ], [ 2, 3 ], [ 3, 1 ], [ 3, 2 ], [ 1, 2, 3 ] ] */
Q6.Travelling sales man problem using brut force algorithm
// Travelling sales man problem using brut force algorithm function calculateDistance(matrix, path) { let totalDistance = 0; for (let i = 0; i < path.length - 1; i++) { totalDistance += matrix[path[i]][path[i + 1]]; } // Return to the starting city totalDistance += matrix[path[path.length - 1]][path[0]]; return totalDistance; } function permute(arr) { const result = []; const helper = (combination, depth) => { if (depth == 4) { result.push([...combination]); return; } for (let item of arr) { if (combination.indexOf(item) < 0) { combination.push(item); helper(combination, depth +1); combination.pop() } } } helper([], 0); return result; } function tsp(matrix) { const cities = Array.from({length: matrix.length}, (_, index) => index) console.log(cities) const permutations = permute(cities); console.log(permutations) let minDistance = Infinity; let bestPath = []; for (let path of permutations) { const distance = calculateDistance(matrix, path); if (distance < minDistance) { minDistance = distance; bestPath = path; } } return { minDistance, bestPath }; } // Example usage: const distanceMatrix = [ [0, 10, 15, 20], [10, 0, 35, 25], [15, 35, 0, 30], [20, 25, 30, 0] ]; const result = tsp(distanceMatrix); console.log(`The shortest distance is: ${result.minDistance}`); console.log(`The best path is: ${result.bestPath}`); /* Initialization: Calculate Distance Explanation totalDistance is initialized to 0. First Iteration (i = 0): From city 0 to city 1: matrix[0][1] is 10. Add 10 to totalDistance, making it 10. Second Iteration (i = 1): From city 1 to city 3: matrix[1][3] is 25. Add 25 to totalDistance, making it 35. Third Iteration (i = 2): From city 3 to city 2: matrix[3][2] is 30. Add 30 to totalDistance, making it 65. Return to Starting City: From city 2 back to city 0: matrix[2][0] is 15. Add 15 to totalDistance, making it 80. Return Total Distance: The function returns 80, which is the total distance of the path [0, 1, 3, 2, 0]. // Output [ 0, 1, 2, 3 ] [ [ 0, 1, 2, 3 ], [ 0, 1, 3, 2 ], [ 0, 2, 1, 3 ], [ 0, 2, 3, 1 ], [ 0, 3, 1, 2 ], [ 0, 3, 2, 1 ], [ 1, 0, 2, 3 ], [ 1, 0, 3, 2 ], [ 1, 2, 0, 3 ], [ 1, 2, 3, 0 ], [ 1, 3, 0, 2 ], [ 1, 3, 2, 0 ], [ 2, 0, 1, 3 ], [ 2, 0, 3, 1 ], [ 2, 1, 0, 3 ], [ 2, 1, 3, 0 ], [ 2, 3, 0, 1 ], [ 2, 3, 1, 0 ], [ 3, 0, 1, 2 ], [ 3, 0, 2, 1 ], [ 3, 1, 0, 2 ], [ 3, 1, 2, 0 ], [ 3, 2, 0, 1 ], [ 3, 2, 1, 0 ] ] The shortest distance is: 80 The best path is: 0,1,3,2 */
Q7. 0/1 knapsack Brut force Problem
// 0/1 knapsack Brut force Problem function knapsackBruteForce(weights, values, capacity) { let n = weights.length; let maxValue = 0; const subsetResult = []; const binaryVals = [0, 1]; // Function to calculate the total weight and value of a subset function calculateSubset(subset) { let totalWeight = 0; let totalValue = 0; for (let i = 0; i < subset.length; i++) { if (subset[i]) { totalWeight += weights[i]; totalValue += values[i]; } } return { totalWeight, totalValue }; } const helper = (combination, depth) => { if (depth == 4) { subsetResult.push([...combination]); return ; } for (let item of binaryVals) { combination.push(item); helper(combination, depth +1); combination.pop() } } helper([], 0); console.log(subsetResult) // Generate all subsets using binary representation for (let subset of subsetResult) { let { totalWeight, totalValue } = calculateSubset(subset); if (totalWeight <= capacity && totalValue > maxValue) { maxValue = totalValue; } } return maxValue; } // Example usage: const weights = [2, 3, 4, 5]; const values = [3, 4, 5, 6]; const capacity = 5; const maxVal = knapsackBruteForce(weights, values, capacity); console.log(`The maximum value in the knapsack is: ${maxVal}`); /* [ [ 0, 0, 0, 0 ], [ 0, 0, 0, 1 ], [ 0, 0, 1, 0 ], [ 0, 0, 1, 1 ], [ 0, 1, 0, 0 ], [ 0, 1, 0, 1 ], [ 0, 1, 1, 0 ], [ 0, 1, 1, 1 ], [ 1, 0, 0, 0 ], [ 1, 0, 0, 1 ], [ 1, 0, 1, 0 ], [ 1, 0, 1, 1 ], [ 1, 1, 0, 0 ], [ 1, 1, 0, 1 ], [ 1, 1, 1, 0 ], [ 1, 1, 1, 1 ] ] The maximum value in the knapsack is: 7 */
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