0494. Target Sum
https://leetcode.com/problems/target-sum
Description
You are given an integer array nums
and an integer target
.
You want to build an expression out of nums by adding one of the symbols '+'
and '-'
before each integer in nums and then concatenate all the integers.
For example, if
nums = [2, 1]
, you can add a'+'
before2
and a'-'
before1
and concatenate them to build the expression"+2-1"
.
Return the number of different expressions that you can build, which evaluates to target
.
Example 1:
**Input:** nums = [1,1,1,1,1], target = 3
**Output:** 5
**Explanation:** There are 5 ways to assign symbols to make the sum of nums be target 3.
-1 + 1 + 1 + 1 + 1 = 3
+1 - 1 + 1 + 1 + 1 = 3
+1 + 1 - 1 + 1 + 1 = 3
+1 + 1 + 1 - 1 + 1 = 3
+1 + 1 + 1 + 1 - 1 = 3
Example 2:
**Input:** nums = [1], target = 1
**Output:** 1
Constraints:
1 <= nums.length <= 20
0 <= nums[i] <= 1000
0 <= sum(nums[i]) <= 1000
-1000 <= target <= 1000
ac1: knapsack dp
class Solution {
public int findTargetSumWays(int[] nums, int S) {
// edge cases
if (nums == null || nums.length == 0) return 0;
int sum = 0;
for (int i : nums) sum += i; // get sum, dp range [-sum, sum]
if (sum < S || -sum > S) return 0; // can't reach S
int[] dp = new int[2*sum + 1];
dp[0+sum] = 1; // when sum == 0, dp is 1
for (int j = 0; j < nums.length; j++) {
int[] dpnew = new int[2*sum + 1];
for (int d = 0; d < dpnew.length; d++) {
if (d - nums[j] >= 0 && d + nums[j] < dp.length) {
dpnew[d - nums[j]] += dp[d];
dpnew[d + nums[j]] += dp[d];
}
}
dp = dpnew;
}
return dp[sum+S];
}
}
/*
knapsack, for each element, you can take it or leave it
for each num, plus/minus it get a result, then cnt++ at that result
*/
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