416. Partition Equal Subset Sum
Classic 0-1 Knapsack probles. Use two dimension array dp[i][j] to represent that end at i-th element can find the sum j.
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class {
public boolean canPartition(int[] nums) {
int sum = 0;
for (int num: nums) {
sum += num;
}
if (sum % 2 == 1) {
return false;
}
int len = nums.length;
boolean[][] dp = new boolean[len+1][sum/2+1];
for (int i = 0; i < len + 1; i++) {
dp[i][0] = true;
}
for (int j = 1; j < sum/2+1; j++) {
dp[0][j] = false;
}
for (int i = 1; i < len + 1; i++) {
for (int j = 1; j < sum/2+1; j++) {
dp[i][j] = dp[i-1][j];
if (j >= nums[i-1]) {
dp[i][j] = dp[i-1][j] || dp[i-1][j-nums[i-1]];
}
}
}
return dp[len][sum/2];
}
}
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we can optimize the space complexity by using one-dimension array dp and update from end to start.
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class {
public boolean canPartition(int[] nums) {
int sum = 0;
for (int num: nums) {
sum += num;
}
if (sum % 2 == 1) {
return false;
}
int len = nums.length;
boolean[] dp = new boolean[sum/2+1];
dp[0] = true;
for (int num: nums) {
for (int j = sum/2; j >= num; j--) {
dp[j] = dp[j] || dp[j-num];
}
}
return dp[sum/2];
}
}
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