Below is the implementation of the target sum:
123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126/** * Find the number of different expressions that evaluate to target. * Recursion with Memoization approach. * * @param {number[]} nums - Array of integers * @param {number} target - Target sum * @return {number} - Number of different expressions */function findTargetSumWays(nums, target) { const memo = new Map(); /** * Helper function to calculate the number of ways to achieve the target sum. * * @param {number} index - Current index in the array * @param {number} sum - Current sum * @return {number} - Number of ways to achieve the target sum */ function calculate(index, sum) { // Base case: reached the end of the array if (index === nums.length) { return sum === target ? 1 : 0; } // Check if we've already computed this subproblem const key = `${index},${sum}`; if (memo.has(key)) { return memo.get(key); } // Add the current number const add = calculate(index + 1, sum + nums[index]); // Subtract the current number const subtract = calculate(index + 1, sum - nums[index]); // Memoize and return const result = add + subtract; memo.set(key, result); return result; } return calculate(0, 0);} /** * Find the number of different expressions that evaluate to target. * Dynamic Programming (0/1 Knapsack) approach. * * @param {number[]} nums - Array of integers * @param {number} target - Target sum * @return {number} - Number of different expressions */function findTargetSumWaysDP(nums, target) { // Calculate the sum of all elements const sum = nums.reduce((acc, num) => acc + num, 0); // Check if it's possible to achieve the target sum if ((sum + target) % 2 !== 0 || Math.abs(target) > sum) { return 0; } const P = (sum + target) / 2; // Initialize DP array const dp = new Array(P + 1).fill(0); dp[0] = 1; for (const num of nums) { for (let j = P; j >= num; j--) { dp[j] += dp[j - num]; } } return dp[P];} /** * Find the number of different expressions that evaluate to target. * 2D Dynamic Programming approach. * * @param {number[]} nums - Array of integers * @param {number} target - Target sum * @return {number} - Number of different expressions */function findTargetSumWays2D(nums, target) { const n = nums.length; // Calculate the sum of all elements const sum = nums.reduce((acc, num) => acc + num, 0); // Check if it's possible to achieve the target sum if (Math.abs(target) > sum) { return 0; } // Initialize 2D DP array const dp = Array(n + 1).fill().map(() => Array(2 * sum + 1).fill(0)); dp[0][sum] = 1; // There's one way to achieve a sum of 0 using 0 elements for (let i = 1; i <= n; i++) { for (let j = 0; j <= 2 * sum; j++) { // Add the current number if (j - nums[i - 1] >= 0) { dp[i][j] += dp[i - 1][j - nums[i - 1]]; } // Subtract the current number if (j + nums[i - 1] <= 2 * sum) { dp[i][j] += dp[i - 1][j + nums[i - 1]]; } } } return dp[n][sum + target];} // Test casesconsole.log(findTargetSumWays([1, 1, 1, 1, 1], 3)); // 5console.log(findTargetSumWays([1], 1)); // 1 console.log(findTargetSumWaysDP([1, 1, 1, 1, 1], 3)); // 5console.log(findTargetSumWaysDP([1], 1)); // 1 console.log(findTargetSumWays2D([1, 1, 1, 1, 1], 3)); // 5console.log(findTargetSumWays2D([1], 1)); // 1
Let's break down the implementation:
Implement the target sum solution in different programming languages.
Below is the implementation of the target sum in different programming languages. Select a language tab to view the corresponding code.
123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126/** * Find the number of different expressions that evaluate to target. * Recursion with Memoization approach. * * @param {number[]} nums - Array of integers * @param {number} target - Target sum * @return {number} - Number of different expressions */function findTargetSumWays(nums, target) { const memo = new Map(); /** * Helper function to calculate the number of ways to achieve the target sum. * * @param {number} index - Current index in the array * @param {number} sum - Current sum * @return {number} - Number of ways to achieve the target sum */ function calculate(index, sum) { // Base case: reached the end of the array if (index === nums.length) { return sum === target ? 1 : 0; } // Check if we've already computed this subproblem const key = `${index},${sum}`; if (memo.has(key)) { return memo.get(key); } // Add the current number const add = calculate(index + 1, sum + nums[index]); // Subtract the current number const subtract = calculate(index + 1, sum - nums[index]); // Memoize and return const result = add + subtract; memo.set(key, result); return result; } return calculate(0, 0);} /** * Find the number of different expressions that evaluate to target. * Dynamic Programming (0/1 Knapsack) approach. * * @param {number[]} nums - Array of integers * @param {number} target - Target sum * @return {number} - Number of different expressions */function findTargetSumWaysDP(nums, target) { // Calculate the sum of all elements const sum = nums.reduce((acc, num) => acc + num, 0); // Check if it's possible to achieve the target sum if ((sum + target) % 2 !== 0 || Math.abs(target) > sum) { return 0; } const P = (sum + target) / 2; // Initialize DP array const dp = new Array(P + 1).fill(0); dp[0] = 1; for (const num of nums) { for (let j = P; j >= num; j--) { dp[j] += dp[j - num]; } } return dp[P];} /** * Find the number of different expressions that evaluate to target. * 2D Dynamic Programming approach. * * @param {number[]} nums - Array of integers * @param {number} target - Target sum * @return {number} - Number of different expressions */function findTargetSumWays2D(nums, target) { const n = nums.length; // Calculate the sum of all elements const sum = nums.reduce((acc, num) => acc + num, 0); // Check if it's possible to achieve the target sum if (Math.abs(target) > sum) { return 0; } // Initialize 2D DP array const dp = Array(n + 1).fill().map(() => Array(2 * sum + 1).fill(0)); dp[0][sum] = 1; // There's one way to achieve a sum of 0 using 0 elements for (let i = 1; i <= n; i++) { for (let j = 0; j <= 2 * sum; j++) { // Add the current number if (j - nums[i - 1] >= 0) { dp[i][j] += dp[i - 1][j - nums[i - 1]]; } // Subtract the current number if (j + nums[i - 1] <= 2 * sum) { dp[i][j] += dp[i - 1][j + nums[i - 1]]; } } } return dp[n][sum + target];} // Test casesconsole.log(findTargetSumWays([1, 1, 1, 1, 1], 3)); // 5console.log(findTargetSumWays([1], 1)); // 1 console.log(findTargetSumWaysDP([1, 1, 1, 1, 1], 3)); // 5console.log(findTargetSumWaysDP([1], 1)); // 1 console.log(findTargetSumWays2D([1, 1, 1, 1, 1], 3)); // 5console.log(findTargetSumWays2D([1], 1)); // 1
First, understand that we need to find the number of different ways to assign '+' and '-' signs to the elements of the array to achieve the target sum.
Use recursion with memoization to explore all possible combinations of '+' and '-' signs, avoiding redundant calculations.
Transform the problem into a subset sum problem and use dynamic programming to find the number of subsets with a specific sum.
Use a 2D dynamic programming array to directly solve the original problem, considering all possible sums at each step.
Handle edge cases such as when it's impossible to achieve the target sum (e.g., when the target is greater than the sum of all elements).
Modify the code to implement an alternative approach and test it with the same examples.
Implement a function that solves the target sum problem using a different approach than shown above.
Handle the case where it's impossible to achieve the target sum (e.g., when the target is greater than the sum of all elements).
Handle the case where the sum of all elements plus the target is odd, making it impossible to find a valid subset sum.
Handle the case where there's only one element in the array.
Below is the implementation of the target sum:
123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126/** * Find the number of different expressions that evaluate to target. * Recursion with Memoization approach. * * @param {number[]} nums - Array of integers * @param {number} target - Target sum * @return {number} - Number of different expressions */function findTargetSumWays(nums, target) { const memo = new Map(); /** * Helper function to calculate the number of ways to achieve the target sum. * * @param {number} index - Current index in the array * @param {number} sum - Current sum * @return {number} - Number of ways to achieve the target sum */ function calculate(index, sum) { // Base case: reached the end of the array if (index === nums.length) { return sum === target ? 1 : 0; } // Check if we've already computed this subproblem const key = `${index},${sum}`; if (memo.has(key)) { return memo.get(key); } // Add the current number const add = calculate(index + 1, sum + nums[index]); // Subtract the current number const subtract = calculate(index + 1, sum - nums[index]); // Memoize and return const result = add + subtract; memo.set(key, result); return result; } return calculate(0, 0);} /** * Find the number of different expressions that evaluate to target. * Dynamic Programming (0/1 Knapsack) approach. * * @param {number[]} nums - Array of integers * @param {number} target - Target sum * @return {number} - Number of different expressions */function findTargetSumWaysDP(nums, target) { // Calculate the sum of all elements const sum = nums.reduce((acc, num) => acc + num, 0); // Check if it's possible to achieve the target sum if ((sum + target) % 2 !== 0 || Math.abs(target) > sum) { return 0; } const P = (sum + target) / 2; // Initialize DP array const dp = new Array(P + 1).fill(0); dp[0] = 1; for (const num of nums) { for (let j = P; j >= num; j--) { dp[j] += dp[j - num]; } } return dp[P];} /** * Find the number of different expressions that evaluate to target. * 2D Dynamic Programming approach. * * @param {number[]} nums - Array of integers * @param {number} target - Target sum * @return {number} - Number of different expressions */function findTargetSumWays2D(nums, target) { const n = nums.length; // Calculate the sum of all elements const sum = nums.reduce((acc, num) => acc + num, 0); // Check if it's possible to achieve the target sum if (Math.abs(target) > sum) { return 0; } // Initialize 2D DP array const dp = Array(n + 1).fill().map(() => Array(2 * sum + 1).fill(0)); dp[0][sum] = 1; // There's one way to achieve a sum of 0 using 0 elements for (let i = 1; i <= n; i++) { for (let j = 0; j <= 2 * sum; j++) { // Add the current number if (j - nums[i - 1] >= 0) { dp[i][j] += dp[i - 1][j - nums[i - 1]]; } // Subtract the current number if (j + nums[i - 1] <= 2 * sum) { dp[i][j] += dp[i - 1][j + nums[i - 1]]; } } } return dp[n][sum + target];} // Test casesconsole.log(findTargetSumWays([1, 1, 1, 1, 1], 3)); // 5console.log(findTargetSumWays([1], 1)); // 1 console.log(findTargetSumWaysDP([1, 1, 1, 1, 1], 3)); // 5console.log(findTargetSumWaysDP([1], 1)); // 1 console.log(findTargetSumWays2D([1, 1, 1, 1, 1], 3)); // 5console.log(findTargetSumWays2D([1], 1)); // 1
Let's break down the implementation:
Implement the target sum solution in different programming languages.
Below is the implementation of the target sum in different programming languages. Select a language tab to view the corresponding code.
123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126/** * Find the number of different expressions that evaluate to target. * Recursion with Memoization approach. * * @param {number[]} nums - Array of integers * @param {number} target - Target sum * @return {number} - Number of different expressions */function findTargetSumWays(nums, target) { const memo = new Map(); /** * Helper function to calculate the number of ways to achieve the target sum. * * @param {number} index - Current index in the array * @param {number} sum - Current sum * @return {number} - Number of ways to achieve the target sum */ function calculate(index, sum) { // Base case: reached the end of the array if (index === nums.length) { return sum === target ? 1 : 0; } // Check if we've already computed this subproblem const key = `${index},${sum}`; if (memo.has(key)) { return memo.get(key); } // Add the current number const add = calculate(index + 1, sum + nums[index]); // Subtract the current number const subtract = calculate(index + 1, sum - nums[index]); // Memoize and return const result = add + subtract; memo.set(key, result); return result; } return calculate(0, 0);} /** * Find the number of different expressions that evaluate to target. * Dynamic Programming (0/1 Knapsack) approach. * * @param {number[]} nums - Array of integers * @param {number} target - Target sum * @return {number} - Number of different expressions */function findTargetSumWaysDP(nums, target) { // Calculate the sum of all elements const sum = nums.reduce((acc, num) => acc + num, 0); // Check if it's possible to achieve the target sum if ((sum + target) % 2 !== 0 || Math.abs(target) > sum) { return 0; } const P = (sum + target) / 2; // Initialize DP array const dp = new Array(P + 1).fill(0); dp[0] = 1; for (const num of nums) { for (let j = P; j >= num; j--) { dp[j] += dp[j - num]; } } return dp[P];} /** * Find the number of different expressions that evaluate to target. * 2D Dynamic Programming approach. * * @param {number[]} nums - Array of integers * @param {number} target - Target sum * @return {number} - Number of different expressions */function findTargetSumWays2D(nums, target) { const n = nums.length; // Calculate the sum of all elements const sum = nums.reduce((acc, num) => acc + num, 0); // Check if it's possible to achieve the target sum if (Math.abs(target) > sum) { return 0; } // Initialize 2D DP array const dp = Array(n + 1).fill().map(() => Array(2 * sum + 1).fill(0)); dp[0][sum] = 1; // There's one way to achieve a sum of 0 using 0 elements for (let i = 1; i <= n; i++) { for (let j = 0; j <= 2 * sum; j++) { // Add the current number if (j - nums[i - 1] >= 0) { dp[i][j] += dp[i - 1][j - nums[i - 1]]; } // Subtract the current number if (j + nums[i - 1] <= 2 * sum) { dp[i][j] += dp[i - 1][j + nums[i - 1]]; } } } return dp[n][sum + target];} // Test casesconsole.log(findTargetSumWays([1, 1, 1, 1, 1], 3)); // 5console.log(findTargetSumWays([1], 1)); // 1 console.log(findTargetSumWaysDP([1, 1, 1, 1, 1], 3)); // 5console.log(findTargetSumWaysDP([1], 1)); // 1 console.log(findTargetSumWays2D([1, 1, 1, 1, 1], 3)); // 5console.log(findTargetSumWays2D([1], 1)); // 1
First, understand that we need to find the number of different ways to assign '+' and '-' signs to the elements of the array to achieve the target sum.
Use recursion with memoization to explore all possible combinations of '+' and '-' signs, avoiding redundant calculations.
Transform the problem into a subset sum problem and use dynamic programming to find the number of subsets with a specific sum.
Use a 2D dynamic programming array to directly solve the original problem, considering all possible sums at each step.
Handle edge cases such as when it's impossible to achieve the target sum (e.g., when the target is greater than the sum of all elements).
Modify the code to implement an alternative approach and test it with the same examples.
Implement a function that solves the target sum problem using a different approach than shown above.
Handle the case where it's impossible to achieve the target sum (e.g., when the target is greater than the sum of all elements).
Handle the case where the sum of all elements plus the target is odd, making it impossible to find a valid subset sum.
Handle the case where there's only one element in the array.