1449. Form Largest Integer With Digits That Add up to Target
Explanation:
To solve this problem, we can use dynamic programming. We create a 1D array dp where dp[i] represents the maximum number we can form with a cost equal to i. We iterate over each digit cost and update the dp array accordingly. Finally, we build the result string by backtracking through the dp array.
- Initialize
dparray with sizetarget + 1and fill it with-1(to indicate it's not possible to form a number with that cost). - Set
dp[0] = 0as the cost of forming number '0' is 0. - Iterate over each digit cost and update
dparray. - Backtrack through the
dparray to build the result string.
Time Complexity: O(target) where target is the given target value. Space Complexity: O(target) for the dp array.
:
class Solution {
public String largestNumber(int[] cost, int target) {
int[] dp = new int[target + 1];
Arrays.fill(dp, -1);
dp[0] = 0;
for (int i = 1; i <= target; i++) {
for (int j = 0; j < 9; j++) {
if (i - cost[j] >= 0 && dp[i - cost[j]] >= 0) {
dp[i] = Math.max(dp[i], dp[i - cost[j]] + 1);
}
}
}
if (dp[target] < 0) {
return "0";
}
StringBuilder sb = new StringBuilder();
for (int i = 8; i >= 0; i--) {
while (target - cost[i] >= 0 && dp[target] == dp[target - cost[i]] + 1) {
sb.append(i + 1);
target -= cost[i];
}
}
return sb.toString();
}
}Code Editor (Testing phase)
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