Summary Ranges

Summary Ranges - Problem

Array Easy

You are given a sorted unique integer array nums.

A range [a,b] is the set of all integers from a to b (inclusive).

Return the smallest sorted list of ranges that cover all the numbers in the array exactly. That is, each element of nums is covered by exactly one of the ranges, and there is no integer x such that x is in one of the ranges but not in nums.

Each range [a,b] in the list should be output as:

"a->b" if a != b
"a" if a == b

Input & Output

Example 1 — Basic Consecutive Ranges

$ Input: nums = [0,1,2,4,5,7]

› Output: ["0->2","4->5","7"]

💡 Note: The ranges are: [0,2] → "0->2", [4,5] → "4->5", [7,7] → "7"

Example 2 — All Single Elements

$ Input: nums = [0,2,3,4,6,8,9]

› Output: ["0","2->4","6","8->9"]

💡 Note: Single element 0, range [2,4], single element 6, range [8,9]

Example 3 — Single Element

$ Input: nums = [1]

› Output: ["1"]

💡 Note: Only one element, so output is just that element as string

Constraints

0 ≤ nums.length ≤ 20
-2³¹ ≤ nums[i] ≤ 2³¹ - 1
All values in nums are unique
nums is sorted in ascending order

Visualization

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Asked in

G Google 15 f Facebook 12 M Microsoft 8

The key insight is to track the start of each consecutive range and detect when the sequence breaks. The optimal approach scans the array once, formatting ranges when gaps are found. Time: O(n), Space: O(1).

Common Approaches

✓ Brute Force - Check Each Element

⏱️ Time: O(n²) Space: O(1)

Iterate through each element and check how many consecutive numbers follow it. Create ranges by examining each position independently.

One Pass Linear Scan

⏱️ Time: O(n) Space: O(1)

Scan the array once, keeping track of the start of the current range. When we find a gap in consecutive numbers, we close the current range and start a new one.

Brute Force - Check Each Element — Algorithm Steps

Step 1: For each element, find the longest consecutive sequence
Step 2: Skip elements already processed
Step 3: Format each range as string

Visualization

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Step-by-Step Walkthrough

Start Check

Begin at each position and check consecutive elements

Find Range

Continue until non-consecutive element found

Format Result

Create range string and add to result

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char** solution(int* nums, int numsSize, int* returnSize) {
    if (numsSize == 0) {
        *returnSize = 0;
        return NULL;
    }
    
    char** result = (char**)malloc(numsSize * sizeof(char*));
    int resultIndex = 0;
    int i = 0;
    
    while (i < numsSize) {
        int start = i;
        // Find end of consecutive sequence
        while (i + 1 < numsSize && nums[i + 1] == nums[i] + 1) {
            i++;
        }
        
        // Format range
        result[resultIndex] = (char*)malloc(50 * sizeof(char));
        if (start == i) {
            sprintf(result[resultIndex], "%d", nums[start]);
        } else {
            sprintf(result[resultIndex], "%d->%d", nums[start], nums[i]);
        }
        
        resultIndex++;
        i++;
    }
    
    *returnSize = resultIndex;
    return result;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array manually
    int nums[1000];
    int numsSize = 0;
    
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        if (strlen(token) > 0 && token[0] != '\n') {
            nums[numsSize++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    int returnSize;
    char** result = solution(nums, numsSize, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("\"%s\"", result[i]);
        if (i < returnSize - 1) printf(",");
        free(result[i]);
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops where outer loop runs n times and inner loop can run up to n times

✓ Linear Growth

Space Complexity

O(1)

Only using variables for tracking, output list not counted

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check Each Element — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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