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RangeSumQueryImmutable.java
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154 lines (117 loc) · 4.32 KB
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package Algorithms.PrefixSum;
import java.util.HashMap;
import java.util.Map;
/**
* @author Srinivas Vadige, srinivas.vadige@gmail.com
* @since 11 July 2025
* @link 303. Range Sum Query - Immutable <a href="https://leetcode.com/problems/range-sum-query-immutable/">LeetCode link</a>
* @description Calculate the sum of the elements of nums between indices left and right inclusive where left <= right.
* @topics Array, Design, Prefix Sum
* @companies Meta, Google, Amazon, Bloomberg, Microsoft, Palantir Technologies, Apple
*/
public class RangeSumQueryImmutable {
public static void main(String[] args) {
int[] nums = {-2, 0, 3,-5, 2,-1};
System.out.println("NumArray using PrefixSum => ");
NumArray numArray = new NumArray(nums);
System.out.println("sumRange(0, 2) => " + numArray.sumRange(0, 2)); // 1
System.out.println("sumRange(2, 5) => " + numArray.sumRange(2, 5)); // -1
System.out.println("sumRange(0, 5) => " + numArray.sumRange(0, 5)); // -3
nums = new int[]{-2, 0, 3,-5, 2,-1};
System.out.println("NumArray2 using PrefixSum => ");
NumArray2 numArray2 = new NumArray2(nums);
System.out.println("sumRange(0, 2) => " + numArray2.sumRange(0, 2)); // 1
System.out.println("sumRange(2, 5) => " + numArray2.sumRange(2, 5)); // -1
System.out.println("sumRange(0, 5) => " + numArray2.sumRange(0, 5)); // -3
}
/**
* @SpaceComplexity O(1)
*/
static class NumArray {
private final int[] prefixSum;
public NumArray(int[] nums) {
this.prefixSum = nums; // modifying the original array
/*
[-2, 0, 3,-5, 2,-1]
[-2,-2, 1,-4,-2,-3] --- prefixSum
*/
for(int i=1; i<nums.length; i++) {
nums[i] += nums[i-1];
}
}
/**
* @TimeComplexity O(1)
* To get the sum of range from left to right --> sum(1,5) = sum(5) - sum(0); 🔥
* sum(i to j) = prefixSum[j] - prefixSum[i - 1]
* so, prefixSum[i-1] = prefixSum[j] - sum(i to j)
* int subtract = prefixSum[i-1]
*/
public int sumRange(int left, int right) {
if(left == 0) {
return prefixSum[right];
}
return prefixSum[right] - prefixSum[left-1];
}
}
/**
* @SpaceComplexity O(n)
*/
static class NumArray2 {
private final int[] prefixSum;
public NumArray2(int[] nums) {
prefixSum = new int[nums.length + 1]; // dummy prefixSum[0]=0;
for(int i = 0; i < nums.length; i++){
prefixSum[i + 1] = prefixSum[i] + nums[i];
}
}
/**
* @TimeComplexity O(1)
*/
public int sumRange(int left, int right) {
return prefixSum[right + 1] - prefixSum[left];
}
}
/**
* @SpaceComplexity O(1)
*/
static class NumArrayUsingBruteForce {
private final int[] nums;
public NumArrayUsingBruteForce(int[] nums) {
this.nums = nums;
}
public int sumRange(int left, int right) {
int res = 0;
for(int i = left; i <= right; i++)
res += nums[i];
return res;
}
}
/**
* @SpaceComplexity O(n^2)
*/
static class NumArrayUsingCaching {
private final Map<Pair, Integer> map = new HashMap<>();
// or use Map<Map.Entry<Integer, Integer>, Integer>> map = new HashMap<>(); and then set key = new java.util.AbstractMap.SimpleEntry<>(1, 2);
static class Pair {
int x, y;
Pair(int x, int y) { this.x = x; this.y = y; }
public boolean equals(Object o) { return o instanceof Pair p && p.x == x && p.y == y; }
public int hashCode() { return 31 * x + y; }
}
/**
* @TimeComplexity O(n^2)
*/
public NumArrayUsingCaching(int[] nums) {
for (int i = 0; i < nums.length; i++) {
int sum = 0;
for (int j = i; j < nums.length; j++) {
sum += nums[j];
map.put(new Pair(i, j), sum);
}
}
}
public int sumRange(int i, int j) {
return map.get(new Pair(i, j));
}
}
}